How do I calculate the peak output voltage of a capacitor differentiator?
You mean like this? 

  

The peak output voltage would just be the peak input voltage.  If it changes fast enough, the output will track it almost exactly, and then decay to smaller values:



Oh wait.  That doesn't take into account a repeating waveform, like this:



After the wave has decayed, when you switch the input again, the output will jump to a lower voltage than the input:



So I think the worst case would be square waves of low frequency, so that they decay completely to zero before switching, and the peak-to-peak output voltage would then be 2× the input peak-to-peak voltage.