How long to charge a super-capacitor with LM317
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.