How about doing the maths...
Energy in a capacitor = 0.5 * C * V^2
Energy in the cap when it's charged to 12V (12V is roughly right- a non-running car, with a 0.3V Schottky diode in series) = 72J * C
Energy left in the cap when the Empeg unit stops running (at 8V?) = 32J * C
so you've got 32Joules times the size of your capacitor to play with.
The Empeg unit takes about 8W (8 joules per second), IIRC.
If your car takes 2 seconds to start, you'll need a 0.5 farad cap. Not unreasonable. Note that there's no slack in there, and some of my assumptions may be wrong, especially the cutoff-voltage one. It's not impossible, though...

STeve