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'll start with the short-term answer, but be aware that it is dangerous.
The relationship you're looking for is $$\frac{dV}{dt} = \frac{i}{C}$$ or, more usefully here, $$\frac{\Delta V}{\Delta t} = \frac{i}{C}$$ and plugging in your numbers gives $$\Delta t = \frac{{\Delta v} \times C}{i} = \frac{2.5 \times 600}{1.5} = 1000 seconds$$
Now for potential problems.
First, and most important, is the matter of cutoff voltage. You'll need to keep at least 3 volts across the LM317 (Section 8.4.2), so let's say you use a 6-volt source on the regulator input. When you get to 2.5 volts, the regulator will not simply cut off. Instead, depending on the unit you've got, it may well keep on supplying current, although probably at a lower level, and may well drive your cap to 5 volts or more. The 3-volt headroom requirement is in no way guaranteed - it is the worst-case number.
Second, you cannot count on the internal regulator to limit your current to 1.5 amps. From Section 7.5 (Electrical characteristics) 1.5 amps is the guaranteed minimum, while typical is 2.2, and there is no stated maximum. You can use a resistor to set the current output, and as long as you set it for 1.5 amps or less you'll be OK, but the cutoff problem remains. Also,
Third, you must worry about thermal effects. If you set the current to 1.5 amps, and use a 6-volt input, at the beginning of the capacitor charge the LM317 will dissipate 9 watts (6 x 1.5), dropping to 5.25 watts when the cap is at full charge. Unless you provide a very good heat sink (capable of keeping your package cool with 10 watts on it) the regulator will go into thermal shutdown, cool off, start conducting and heat up, cool down, etc.
Best Answer
Let's look at a simple analysis first. The basic equation relating current and voltage is $$\frac{dV}{dt} = \frac{i}{C} $$ where i is in amps and C is in farads. For a current of .03 amps and 58.33 farads $$\frac{dV}{dt} = 5.14 x 10^{-4} $$ so in one hour $$\Delta V = 1.85\text{ volts}$$ Since 14.4 minus 11.5 equals 2.9 volts, your ultra caps will last about $$\Delta t = \frac{2.9}{1.85} = 1.57\text{ hours}$$
This is so short a period that you simply don't need to worry about self-discharge currents.
EDIT - Along the same lines, you need to take a look at the load on a car battery. The big-ticket item is starting. Starting loads are typically 100 to 200 amps.
For a current of 100 amps (and this is best-case, remember. Starting in cold temperatures takes rather more.) we find $$\frac{dV}{dt} = 1.71 $$ and for a 2.9 volt drop, $$\Delta t = \frac{2.9}{1.71} = 1.69\text{ seconds}$$ For 200 amps, of course, the time is half that, or about 0.85 seconds.
So you might want to reconsider using caps (ultra or otherwise) for a car battery replacement. Also, just for fun, you need to check the spec sheet for the maximum current which can be drawn from an ultracap without damaging it.