Piezo buzzer current consumption
If you're running it at resonance the current won't have a lot to do with the Xc because the mechanical energy is being changed back into electrical energy as the ceramic and backing disc flexes, like a tank (LC) circuit where energy circulates back and forth between the L and the C. 

It will have more to do with the non-useful losses (series equivalent resistance, mechanical losses at the flexure, mechanical losses in the materials, damping losses in the sides of the Helmholtz cavity etc.) and the intentional transfer of power to the air (or whatever medium it is). 

Then there are losses in the driving circuit. 

Measuring it under representative conditions may be your best bet to get an approximate answer quickly, though I'm sure it could be modeled in a multiphysics package like Comsol given the time and inclination.