Why does a 10A fuse blow at 10A and 5V or 10A and 500V?
What blows the fuse is really the power wasted at it.   \$p(t)=u(t)\cdot i(t)\$.

But the voltage here is not the supply voltage of the circuit, but the voltage drop across the fuse, which is determined by the current passed through it.  \$p(t)=R\cdot i^2(t)\$. R is constant, so it really depends only on the current.

In typical setting, the resistance of the fuse will be negligible compared to resistance of the load, so the current running through it will be determined solely by the load resistance (or impedance in general).

In case of short circuit, there will be no load impedance and current will be limited only by the fuse's filament. In such case, supply with higher voltage will make it blow faster. (\$p(t)=\frac{u^2(t)}{R}\$)