The TLC5940 requires a minimum headroom (anode voltage applied to LED) of about 0.7 Volts greater than the LED's Vfwd for driving 60 mA, and 1.2 Volts for 120 mA.
If the headroom is lower than this, the channel is detected as an open LED. Actually, "open" is detected at 0.4 Volts or lower headroom, but that's a minor detail.
In discussions on TI's E2E forum, it has been confirmed from time to time that individual channels (LEDs) can be sourced by differing voltages, as long as the headroom requirement is met.
Another suggested method of reducing the surplus voltage across the TLC5940 driving transistors, is to use an external resistor for each LED, calculated to reduce the maximum current (if the TLC5940 were replaced with a short circuit to ground), to a bit over 10 mA more than the intended LED drive current. That way, the excess voltage is dissipated across each resistor, rather than across the LED driver IC.
Direct answer to the question
The direct answer to your question, assuming you intend to just connect the capacitor to the LED with a series resistor is no time at all. That is because a white LED takes more than 2.7 V to light. Check its datasheet. These things usually need a bit over 3 V.
There are two options. The simplest is to use a LED with a lower forward drop. Let's say you try this with a red LED that has a 1.8 V drop at 20 mA. That means at full charge, there will be 2.7V - 1.8V = 900 mV accross the resistor. If you want the maximum brightness at full charge, which we are saying is 20 mA, then you need a 900mV / 20mA = 45 Ω resistor. Let's pick the common nominal value of 47 Ω.
Now that we have a capacitance and resistance we can compute the time constant, which is 150F x 47Ω = 7050 s = 118 minutes = 2 hours. At full charge, the LED will be nearly at full brightness, which will then decay slowly. There is no fixed limit at which it will suddenly go out, so we have to pick something. Let's say 5 mA is dim enough to be considered not usefully lit anymore in your application. The voltage accross the resistor will be 47Ω x 5mA = 240mV. Using the first approximation of the LED having constant voltage accross it, that means the capacitor voltage is 2 V.
The question is now how long does it take to decay from 2.7 V to 2.0 V at a 2 hour time constant. That is .3 time constants, or 2100 seconds, or 35 minutes. The actual value will be a bit longer due to the LED having some effective series resistance too and therefore increasing the time constant.
A better way
The above tries to answer your question, but is not useful for a flashlight. For a flashlight you want to keep the light at close to the full brightness for as long as possible. That can be done with a switching power supply, which transfer Watts in to Watts out plus some loss but at different combinations of voltage and current. We therefore look at the total energy available and required and not worry about specific volts and amps too much.
The energy in a capacitor is:
$$E = \frac{C \times V^2}{2}$$
When C is in Farads, V in Volts, then E is in Joules.
$$\frac{150F * (2.7V)^2}{2} = 547 J$$
The switching power supply will need some minimum voltage to work with. Let's say it can operate down to 1 V. That represents some energy left in the cap the circuit can't extract:
$$\frac{150F * (1.0V)^2}{2} = 75 J$$
The total available to the switching power suppy is therefore 547 J - 75 J = 470 J. Due to the low voltages, the losses in the switching power supply will be quite high. Let's say that in the end only 1/2 the available energy gets delivered to the LED. That leaves us with 236 J to light the LED.
Now we need to see how much power the LED needs. Let's go back to your original white LED and pick some numbers. Let's say it needs 3.5 V at 20 mA to shine nicely. That's 3.5V * 20 mA = 70 mW. (236 J)/(70 mW) = 3370 seconds, or 56 minutes. At the end of that, the light would go dead rather quickly, but you will have fairly steady brightness up until then.
Best Answer
Less than 8 Seconds
calculations below
THE DRIVER
LED DRIVER FEATURES (Datasheet page 1)
LED DRIVER RECOMMENDED OPERATING CONDITIONS (datasheet page 2)
The LED driver's datasheet examples use two 0.55 Farad Super Caps.
What makes the flash driver so awesome is how it charges, from a 2-5.5V battery, and balances 2 super caps, and gets the flash ready quickly for the next flash.
The LED driver charges two super caps not one.
The LEDs
You are not powering a flash LED, so a boost regulator would do the job.
Given what the datasheet says and the LED and Super Cap you selected you want to light up two LEDs in torch mode.
The LED you selected is not a Camera Flash LED.
The LED datasheet says the applications are:
At 200mA max the LEDs will output about 100 lumens each.
If I assume correctly the number of LEDs does not matter. You just want about 100-200 lumens for as long as a 1F Super Cap will allow.
The Super Cap
You are using one 1 farad super capacitor.
A single digit Farad supercap is used for quick charge discharge, not a storage cell.
A supercap energy cell is in the thousands of Farads.
As an energy cell a supercap is used to bridge power gaps lasting from a few seconds to a few minutes. Like to supply energy between power failure and backup generator starting.
Unless that awesome chip came with some free super caps I doubt you will want to use a supercap in place of a battery. The specific energy (capacity) of a supercap is 10-50 times less than a Li-ion battery. And they cost a lot more than a battery.
Because the cap you specified is 5V, it is not a super cap. But that does not matter because you are charging it to 3.0v, so it's just a chemistry technicality. Some of the most recent super caps are now 3.0v.
Max super cap cell voltage (of reasonably priced and lower capacity) is 2.75V, typically 2.3 to 2.6V max, which is less than the minimum forward voltage of your 2.8 LED.
Conclusion
Given your selection of driver and non-flash LED it appears you want some light generated by LED(s) powered by a Super Cap.
It appears you do NOT need the features of the flash driver other than to charge the Super Cap. And that may not be necessary.
The driver only lights up one LED at a time, and charges one cap at a time.
The 8 seconds still applies to the selected LED driver.
Calculations
Single LED
Driven by Super Cap and Resistor
Given the assumptions, the number of LEDs does not matter. The selected driver charges and discharges one LED at a time.
Let's say you power a single LED with a 2V forward voltage max. The LED you picked has a current of 200mA max. To power it directly with a resistor, at 200mA, 3V you need a 5.6Ω resistor. 1F x 5.6Ω = 5.6 seconds TC.
But the cap will not discharge like a battery.
With 400mV across the resistor (200mA x 5.6Ω) and 2v across the LED the cap voltage can drop from 3 to 2.6 or 600mV before the LED stops conducting.
600mV/3.0 = 0.2
0.2v x 5.6 seconds =1.12
It will take about 1.12 seconds to drop below the 2V forward voltage.
Why We Use an LED Driver, Not a Resistor
This applies just as well to battery power as it does a Super Cap.
The LED driver uses a switching regulator to driver the LED from the Super Cap. So we just need to calculate the energy provided by the cap, consumed by the LED, and how long it takes to discharge the cap at 200mA.
Energy capacity of SC: 1F* 3v2 / 2 = 4.5 Joules
The LED 2.8v * 0.2A = 0.560V
4.5 J / .560 = 8 seconds
With the selected LED driver, in torch mode, it would be 4 seconds for LEDA and 4 seconds for LEDB (withn two 0.5F caps) which could be lit up consecutively giving 8 seconds.
The above assumes an ideal LED driver with 100% efficiency. So it will be less than 8 seconds.
To power the LEDs with a 3.6v 1000mA hr capacity camera battery would power the LED for ____?
You now have the formulas to calculate that for your self. Hint: Much much longer. You can now also calculate the flash discharge if you desire.
If it is true you just want to light up some LEDs with a super cap now you can see why it makes no sense to charge a super cap with a battery unless you need the short (<400mS) 2A flash discharge.
Why the Driver does not give you the option to power the LEDs with the battery, I do not know. That's what I may have done if it were my project.