By translation assume that the center of mass of $\Omega$ is $0$. Let $A$ normalize $\Omega$ and let

$$\tilde{v}(x)=(\det A)^{-2/n}v(Ax).$$

Then

$$M\tilde{v}(\tilde{\Omega})=(\det A)^{-1}Mv(\Omega)$$

with $B_{1}\subset\tilde{\Omega}\subset B_{C(n)}$.

The maximum of $|\tilde{v}|$ is achieved at some point $\tilde{x}\in\tilde{\Omega}$. Let $K$ be the function whose graph is the cone generated by $(\tilde{x},\tilde{v}(x))$ and $\partial B_{C(n)}$. By convexity,

$$M\tilde{v}(\tilde{\Omega})\geq|\nabla K(\tilde{x})|.$$

Since $\nabla K(\tilde{x})$ is a ball of radius at least $c(n)|\min_{\tilde{\Omega}}\tilde{v}|$, we have

$$|\nabla K(\tilde{x})|\geq c(n)|\min_{\tilde{\Omega}}\tilde{v}|^{n}\geq c(n)|\det A|^{-2}|\min_{\Omega}v|^{n}.$$

Finally, $|\det A|\leq C(n)|\Omega|$ so the conclusion follows. ∎