How does ohm's law for capacitors affect this scenario?
The fact of the matter is that both ends of C1 are connected to the same equivalent node (shorted). At steady-state, there is certainly no current flow, because there is no difference in potential on either side of C1. 

To make this more clear: You cited i = C dv/dt. "dv" means difference/change in voltage. But there cannot be a difference in voltage if both ends of C1 are the same equivalent node with 0Ω between.

If there were a charge on C1 before both ends were shorted, then upon shorting, the current would flow (but with theoretical 0 Ω wire, the drain would would be instantaneous / infinite current).