My first attempt mistook the question, let me try again.

By construction, the $2n\times 2n$ complex matrix $M$ is <A HREF="https://en.wikipedia.org/wiki/Skew-Hermitian_matrix">skew-Hermitian</A> and <A HREF="https://en.wikipedia.org/wiki/Hamiltonian_matrix">Hamiltonian</A>, which means that it has the $n\times n$ block decomposition
$$M=\begin{pmatrix}
A&B\\ C&-A^T\end{pmatrix},\;\;\text{with}\;\;A=-A^\ast,\;\;B=B^T=-C^\ast=-\bar{C}.$$
Here $M^T$ denotes the transpose, $\bar{M}$ the complex conjugate, and $M^\ast$ the conjugate transpose.  

Let me assume $\det A\neq 0$. Then <A HREF="https://en.wikipedia.org/wiki/Schur_complement">Schur's determinant identity</A> gives

$$\det M=\det(-AA^T-ACA^{-1}B)=\det(A\bar{A}+A\bar{B}A^{-1}B)$$
$$\qquad=\det(A\bar{A})\det(1+\bar{A}^{-1}\bar{B}A^{-1}B)$$
$$\qquad=|\det A|^2\det(1+\bar{X}X),\;\;\text{with}\;\;X=A^{-1}B.$$

It remains to prove that for any complex matrix $X$, the determinant $\det(1+\bar{X}X)\geq 0$, which requires some further steps. (I have asked at <A HREF="https://math.stackexchange.com/q/4039880/87355">MSE.</A>)