I was wondering whether there was a way to bound the smallest eigenvalue of $M^{\top} D M$ when $M$ is skew-symmetric and $D$ is a positive diagonal matrix.
In particular, I was hoping that $M^{\top} D M$ would be positive definite, but it seems that this might only be the case if $M$ is full rank (since $D$ is not just a positive multiple of the diagonal).
If $D$ is positive diagonal (i.e. diagonal with strictly positive diagonal entries), then for any (real) matrix $M$, $M^TDM$ will be positive semidefinite with rank equal to that of $M$. If $M$ has linearly independent columns, this means that $M^TDM$ will be positive definite.
To see that it $M^TDM$ is positive semidefinite, it suffices to note that for any vector $x$, $$ x^T(M^TDM)x = (Mx)^TD(Mx) \geq 0. $$ To see that the rank is the same as that of $M$, note that $$ \operatorname{rank}(M^TDM) = \operatorname{rank}([D^{1/2}M]^T[D^{1/2}M]) = \operatorname{rank}(D^{1/2}M) = \operatorname{rank}(M). $$ For your particular case, we can also get a bound on the smallest eigenvalue when the size of $M$ is even. Suppose that $M$ and $D$ are square even size. The smallest singular value of $M$ is the smallest absolute value among the eigenvalues of $M$, and the smallest singular value of $D$ is simply the smallest value on the diagonal. We have $$ \sigma_{\min}(M^TDM) \geq \sigma_{\min}(M^T)\sigma_{\min}(D)\sigma_\min(M) = \sigma_\min(M)^2 \sigma_\min(D). $$ On the other hand, $M^TDM$ is positive semidefinite, which means that its singular values are equal to its eigenvalues.