Why is output voltage being neglected when calculating power supplied by this current source?
Since the current source is in parallel with p3, its voltage is the same as p3's: 8 volts. That is, 8 volts is "the voltage that it is outputting as a source".  I'm not sure what other voltage you're talking about.

It's true that a real-world current source will include circuitry which will dissipate more power than is delivered to the load, but since there is no way to know what that is, even in principle, there is no way to calculate it and it is ignored. Just as the voltage source will, in real life, dissipate more power than it delivers to the load, that power is also ignored for the purposes of this sort of problem.