How long would a 150 Farad Capacitor light up an LED
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 &Omega; resistor.  Let's pick the common nominal value of 47 &Omega;.

Now that we have a capacitance and resistance we can compute the time constant, which is 150F x 47&Omega; = 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 47&Omega; 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.