The impedance related bit is easy, the complicated bit is what actually happens within the iPod.
Here we go with a whole string of assumptions, all of which we'll assume are true for a first pass aproximation of the problem.
If the sensitivity (dB/mW) of the earphones in question is the same then they require the same input power to produce the same output SPL.
Because of the impedance difference this means that the speakers are operating at different voltages and currents, for example -
@20mW input, the 19 Ohm A8's are operating at an (rms) voltage of 0.616V and a current of 32.4mA, while the 32 Ohm Apple earbuds have 0.8V and 0.025mA respectively.
The A8's are operating at a higher current (albeit a lower voltage) than the Earbuds. What happens next depends upon the iPod internals. Assuming the iPod doesn't have clever voltage supply control schemes i.e. the headphone amplifier voltage rail is a fixed value, then it is all down to the topology of the headphone amplifier.
If the amp. is a linear type, then the current difference above will result in a reduction in battery runtime when running with A8's. The major power hungry components in the iPod are the HD (if present), the backlight and the audio out, therefore although the A8 current is ~ 30% higher than the Earbud the runtime reduction will be less than this when including the fixed drain of the other components.
If however the amp. is a switching amplifier, then the efficiency will be very high (it is packets of energy being switched into the load rather than linear voltage regulation) and the difference in runtime due to the earphone impedance will be negligable .
So it seems the answer is some where between hardly noticeable and less than 30% when listening at the same volume.
That's that cleared up then
Generally speaking, you aren't learning much if your lips are moving.