The correct answer is that you'd do a complete circuit analysis of the operating points to determine what the voltages are at most nodes -- a subject that would be a non-trivial assignment for the EE student, and more than what I could post here.
That said, if you're observant (and have analyzed enough circuits to have a good "feel" for the way things work), this amp has a direct-coupled output, so it needs to be biased such that the DC offset at Vout is very close to zero or it wouldn't be very nice to speakers. The feedback resistors R8 and R6 forms a voltage divider for AC signals but not for DC due to the C4 blocking capacitor, and there is very little current flowing through R8 because it goes into the emitter junction of Q4 where most of the current is regulated by the CCS (formed by Q5 and Q6), so there is very little voltage drop across R8. This means that the emitter DC voltage of Q4 would be near zero, and therefore the voltage at C4 is also near zero.
Under ideal conditions C4 should "see" just about zero DC volts across it, but if there is a failure (such as a short across the C-E junction in either of the output transistors), then the C4 voltage could go near one of the supply rails. It is thus prudent to choose C4's voltage rating such that it is at least as high as one of the rails.
