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.
I believe that your best bet is to use a dedicated RTC chip for the clock function. These can connect to your microcontroller via an I2C bus and thus take very few I.O pins to support. If fact if you have already deployed the the I2C bus for other purposes then supporting the RTC chip takes no additional pins.
The reason for using an RTC chip over trying to backup your microcontroller is the difference in achievable standby operating current. RTC chips are optimized for extremely low power operation, particularly if you deploy them with their own attached 32.768 KHz crystal.
I have also found that the external RTC chips with crystal can achieve far better long term accuracy. In fact, if you select the right RTC chip you can use internal registers to trim in the oscillator to get even less clock drift over time.
The use of super capacitors is OK but in my experience, with a very low power RTC chip, that it may very well be cheaper and take less board space to use a small coin cell to power the RTC chip. A soldered in battery holder for a BR1225 coin cell will let you slide in the cell and take board space less than .5x.5". Such cell with 48mAH capacity will keep the RTC supported for years. Even better, with the proper RTC chip selection, is that the chip provides a separate pin for the battery connection so it provides all the regular VCC rail to battery switch over circuitry on die.
Best Answer
Hold up time is
T= \$\frac{C(V_s - V_f)}{I}\$
where I is the current, C is the capacitance, Vs is initial voltage on the capacitor, Vf is final voltage on the capacitor (perhaps the minimum voltage at which the system will work).
That's for an ideal capacitor. If the capacitor has significant internal resistance the voltage will drop an additional amount I*R, so the hold up time will be reduced. For a non-ideal capacitor, also adjust I to add the internal leakage current.
If you're trying to hold up a RPi long enough for an orderly shutdown I think you're going to require a very large supercapacitor with low internal resistance or a battery.