In Appendix B of https://arxiv.org/abs/2103.10776 (J. Phys. A: Math. Theor. 54, 375201 (2021)), I derived a transformation of the block Vandermonde determinant above to a Hankel matrix, which in my case dramatically simplified the problem. This is a block-matrix generalization of the well known "Vandermonde factorization of a Hankel matrix":
Let $\mathbf V_{\!\vec{x}}$ be a general Vandermonde matrix \begin{equation} \mathbf V_{\!\vec{x}} \equiv \left[\vphantom{A_{2}^{2}}x_{\mu}^{j}\right]_{\mu=1,\,j=0}^{M\hspace{1.5ex}M-1}\label{eq:V_x}\tag{1} \end{equation} and let \begin{equation} P_{\vec{x}}(x)=\prod_{\mu=1}^{M}(x-x_{\mu})=\sum_{n=0}^{M}b_{n}x^{n},\label{eq:P_x(x)}\tag{2} \end{equation} be the related characteristic polynomial in $x$, where $b_{M}=1$ by construction. If we further let \begin{align} \mathbf D_{\vec{x}} & \equiv \left[\frac{\delta_{\mu\nu}}{P_{\vec{x}}'(x_{\mu})}\right]_{\mu,\nu=1}^{M}, & \mathbf H_{\vec{x}}^{-1} & \equiv\left[\vphantom{A_{2}^{2}}b_{i+j+1}\right]_{i,j=0}^{M-1},\tag{3} \end{align} then \begin{align} \mathbf V^T_{\!\vec{x}}\,\mathbf D_{\vec{x}}\mathbf V_{\!\vec{x}} & = \mathbf H_{\vec{x}}, & {\det}^{2}\,\mathbf V_{\!\vec{x}}\det\mathbf D_{\vec{x}} & =1.\label{eq:Vandermonde_Hankel_Identity}\tag{4} \end{align} While $\mathbf H_{\vec{x}}^{-1}$ is upper anti-triangular, $\mathbf H_{\vec{x}}$ itself is lower anti-triangular, e.g. for $M=4$, \begin{equation} \mathbf H_{\vec{x}}^{-1}=\begin{pmatrix}b_{1} & b_{2} & b_{3} & 1\\ b_{2} & b_{3} & 1 & 0\\ b_{3} & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix}\quad\Rightarrow\quad\mathbf H_{\vec{x}}=\begin{pmatrix}0 & 0 & 0 & 1\\ 0 & 0 & 1 & \tilde{b}_{5}\\ 0 & 1 & \tilde{b}_{5} & \tilde{b}_{6}\\ 1 & \tilde{b}_{5} & \tilde{b}_{6} & \tilde{b}_{7} \end{pmatrix},\label{eq:H_x_example}\tag{5} \end{equation} with $\tilde{b}_{n}=\sum_{\mu}x_{\mu}^{n-1}/P_{\vec{x}}'(x_{\mu})$.
As a side note, both $\tilde{b}_{n}=s_{(n-M)}(\boldsymbol{x})$ and $b_{n}=(-1)^{n}s_{\boldsymbol{1}_{M-n}}(\boldsymbol{x})$ are Schur polynomials.
We now turn to the block matrix version:
Let $\boldsymbol{\mathcal{V}}_{\vec{g},\vec{x}}$ be the generalized $1\times B$ block Vandermonde matrix with $M\times N$ blocks \begin{equation} \boldsymbol{\mathcal{V}}_{\vec{g},\vec{x}} \equiv \left[\vphantom{A_{2}^{2}}g_{\mu}^{b}\,x_{\mu}^{n}\right]_{b=0;\,\,\mu=1,\,n=0}^{B-1\,M\hspace{2ex}N-1} = \left[\vphantom{A_{2}^{2}}\mathbf G^{b}\,\mathbf V_{\!\vec{x}}\right]_{b=0}^{B-1},\tag{6} \end{equation} with \begin{equation} \mathbf V_{\!\vec{x}} \equiv \left[\vphantom{A_{2}^{2}}x_{\mu}^{n}\right]_{\mu=1,\,n=0}^{M\hspace{1.5ex}N-1},\qquad \mathbf G \equiv \left[\vphantom{A_{2}^{2}}\delta_{\mu\nu}\,g_{\mu}\right]_{\mu,\nu=1}^{M}.\tag{7} \end{equation} As an example, for $M=4$, $B=3$ and $N=3$ we have \begin{equation} \boldsymbol{\mathcal{V}}_{\vec{g},\vec{x}} = \left[\begin{array}{ccc|ccc|ccc} 1 & x_{1} & x_{1}^{2} & g_{1} & g_{1}x_{1} & g_{1}x_{1}^{2} & g_{1}^{2} & g_{1}^{2}x_{1} & g_{1}^{2}x_{1}^{2}\\ 1 & x_{2} & x_{2}^{2} & g_{2} & g_{2}x_{2} & g_{2}x_{2}^{2} & g_{2}^{2} & g_{2}^{2}x_{2} & g_{2}^{2}x_{2}^{2}\\ 1 & x_{3} & x_{3}^{2} & g_{3} & g_{3}x_{3} & g_{3}x_{3}^{2} & g_{3}^{2} & g_{3}^{2}x_{3} & g_{3}^{2}x_{3}^{2}\\ 1 & x_{4} & x_{4}^{2} & g_{4} & g_{4}x_{4} & g_{4}x_{4}^{2} & g_{4}^{2} & g_{4}^{2}x_{4} & g_{4}^{2}x_{4}^{2} \end{array}\right]. \end{equation} Furthermore, let $\mathbf D$ be an arbitrary $M\times M$ diagonal matrix. Then, the $B\times B$ block Hankel matrix with $N\times N$ blocks \begin{equation} \boldsymbol{\mathcal{H}}_{\vec{g},\vec{x}} \equiv \left[\sum_{\mu=1}^{M}d_{\mu}g_{\mu}^{a+b}\,x_{\mu}^{m+n}\right]_{a,b=0;\,m,n=0}^{B-1\;\;N-1} = \left[\mathbf V^T_{\!\vec{x}}\,\mathbf D\,\mathbf G^{a+b}\,\mathbf V_{\!\vec{x}}\right]_{a,b=0}^{B-1}\tag{8} \end{equation} trivially fulfills the identity \begin{equation} \boldsymbol{\mathcal{H}}_{\vec{g},\vec{x}} = \boldsymbol{\mathcal{V}}^T_{\vec{g},\vec{x}} \, \mathbf D \, \boldsymbol{\mathcal{V}}_{\vec{g},\vec{x}}.\tag{9} \end{equation} Note that the upper left block $\hat{\mathbf H}_{0} \equiv (\boldsymbol{\mathcal{H}}_{\vec{g},\vec{x}})_{0,0} = \mathbf V^T_{\!\vec{x}}\,\mathbf D\,\mathbf V_{\!\vec{x}}$ is free of $g_{\mu}$ and is therefore a usual Vandermonde product in $x_{\mu}$. Consequently, if the matrix $\mathbf D$ is set to the diagonal matrix $\mathbf D_{\vec{x}}$ containing the reciprocal first derivatives of the characteristic polynomial $P_{\vec{x}}(x)=\prod_{\mu=1}^M(x-x_\mu)$, \begin{equation} \mathbf D_{\vec{x}} \equiv \left[\frac{\delta_{\mu\nu}}{P_{\vec{x}}'(x_{\mu})}\right]_{\mu,\nu=1}^{M},\tag{10} \end{equation} and if additionally $M\geq2N$, then $\hat{\mathbf H}_{0}=\mathbf 0$ vanishes identically.
For the problem above, $B=2$ and $N=M/2$, such that \begin{equation} \mathbf A^T \mathbf D_{\vec{x}}\mathbf A = \boldsymbol{\mathcal{H}}_{\vec{g},\vec{x}} = \begin{bmatrix} \mathbf 0 & \hat{\mathbf H}_{1} \\ \hat{\mathbf H}_{1} & \hat{\mathbf H}_{2} \end{bmatrix}.\tag{11} \end{equation}
The desired determinant $\det \mathbf A$ therefore fulfills \begin{equation} \det(\mathbf A^T \mathbf D_{\vec{x}} \mathbf A) = {\det}^{2}\,\mathbf A\det\mathbf D_{\vec{x}} = \det \boldsymbol{\mathcal{H}}_{\vec{g},\vec{x}} = {\det}^{2}\,\hat{\mathbf H}_{1}.\tag{12} \end{equation}
For further details, see https://arxiv.org/abs/2103.10776.
