The V = IR equation relates the voltage _difference_ across a resistor to it's resistance and the amount of current flowing though it. So a super conductor in a circuit will have 0 drop in voltage across it, while still having an "absolute" voltage (or rather voltage difference between it and a ground).
In theory this it's unrelated to the amount of current passing through the superconductor, but in reality a material will stop acting like a superconductor if you try to push too much amperage through it.
AFAIK the resistance is truly 0. However, there is a critical magnetic field, which if exceeded will destroy the superconducting state. Therefore a current large enough to generate this field will also destroy the superconductivity.
This is correct. Resistance in a superconductor is exactly zero. However if the current reaches a critical level the superconductor will ["quench"](https://en.wikipedia.org/wiki/Superconducting_magnet#Magnet_...) and rapidly drop out of the superconducting state. When that happens, it regains its normal resistance and because there is still a lot of current going through it, the material will heat up rapidly, thus pushing it away from superconductivity even further. This can damage or even destroy the coil if the energies involved are large enough.
In theory this it's unrelated to the amount of current passing through the superconductor, but in reality a material will stop acting like a superconductor if you try to push too much amperage through it.