My first attempt mistook the question, let me try again.
By construction, the $2n\times 2n$ complex matrix $M$ is skew-Hermitian and Hamiltonian, 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 Schur's determinant identity 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 MSE.)