You use Ohm's law:
I=V/R
(I is current, V is voltage, R is resistance)
For our situation, we would reorder it to be R=V/I. So, the equation would look like this.
(14.4V-[Vf of LED])
------------------
[current of LED at Vf]
That will give you the resistor value in Ohms. I'd bump it up just a tad as well. For my 3.2V Vf LEDs, with 20mA of current, I use 620Ohm resistors.
As far as someone doing it, I've considered doing it for about $10 a switch (plus shipping, you provide the switches) but I've never opened Pontiac switches. I'd assume they're similar to the ones in the Impala/Monte, but I'm still learning those.