Well, Binet–Cauchy plus Lindström–Gessel–Viennot work indeed.

Let $i,j$ vary from 0 to $m-1$, and $k$ vary from $0$ to $m+n-1$. Denote also $k^*=m+n-1-k$, so $k^*$ also varies from 0 to $m+n-1$. Consider the $m\times (m+n)$ matrices $A=(a_{i,k})$ and $B=(b_{j,k})$ defined by $$a_{i,k}=(-1)^k{-i-1\choose k}={i+k\choose i}\\ b_{j,k}=(-1)^{k^*}{-j-1\choose k^*}={j+k^*\choose j}.$$ Note that for fixed $i$, $a_{i,k}$ as a function of $k$ is a polynomial of degree $i$ with leading coefficient $1/i!$; and so is $(-1)^ib_{i,k}$. Thus the minors of $A$, $B$ indexed by $\mathbf{J}=\{j_1<j_2<\ldots<j_m\}$ are equal to $V(\mathbf{J})/V(\mathbf{I})$ and $(-1)^{m\choose 2}V(\mathbf{J})/V(\mathbf{I})$ respectively. Thus by Binet–Cauchy we have $$ \sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2=(-1)^{m\choose 2}\det AB^t. $$ As for $AB^t$, its elements $c_{ij}$ are equal to $$ c_{ij}=\sum_{k=0}^{m+n-1}a_{i,k}b_{j,k}=(-1)^{m+n-1}\sum_{k=0}^{m+n-1} {-i-1\choose k}{-j-1\choose k^*}\\=(-1)^{m+n-1}{-i-j-2\choose m+n-1}= {i+j+m+n\choose m+n-1}=P((0,-i),(m+n-1,j+1)), $$ where, for $u,v\in \mathbb{Z}^2$, $P(u,v)$ is the number of lattice paths from $u$ to $v$ (paths go up and right). So, $\det AB^t$ has Lindström–Gessel–Viennot interpretation. The collection of $m$ non-intersecting lattice paths from the points $u_i:=(0,-i)$ to the points $v_j:=(m+n-1,j+1)$ of course must join $u_i$ with $v_{m-1-i}$. This, in particular, gets $(-1)^{m\choose 2}$ factor out. Also, the path from $v_i$ must start with a horizontal part of length at least $i$; analogously for finishing paths. So, counting these paths is equivalent to counting the non-intersecting paths which go from $(i,-i)$ to $(n+i, m-i)$. These paths correspond to Young diagrams in $m\times n$ boards, and the condition of them being non-intersecting is equivalent to corresponding diagrams be $m$ consecutive sections of a 3d Young diagram in the $m\times m\times n$ box. (This last argument is a standard proof of MacMahon formula).

Since Binet–Cauchy itself has interpretation in terms of Lindström–Gessel–Viennot, you may rephrase this argument more combinatorially, if you prefer. I expect that what is obtained is exactly the answer outlined by Richard Stanley.

Well, Binet–Cauchy plus Lindström–Gessel–Viennot work indeed.

Let $i,j$ vary from 0 to $m-1$, and $k$ vary from $0$ to $m+n-1$. Denote also $k^*=m+n-1-k$, so $k^*$ also varies from 0 to $m+n-1$. Consider the $m\times (m+n)$ matrices $A=(a_{i,k})$ and $B=(b_{j,k})$ defined by $$a_{i,k}=(-1)^k{-i-1\choose k}={i+k\choose i}\\ b_{j,k}=(-1)^{k^*}{-j-1\choose k^*}={j+k^*\choose j}.$$ Note that for fixed $i$, $a_{i,k}$ as a function of $k$ is a polynomial of degree $i$ with leading coefficient $1/i!$; and so is $(-1)^ib_{i,k}$. Thus the minors of $A$, $B$ indexed by $\mathbf{J}=\{j_1<j_2<\ldots<j_m\}$ are equal to $V(\mathbf{J})/V(\mathbf{I})$ and $(-1)^{m\choose 2}V(\mathbf{J})/V(\mathbf{I})$ respectively. Thus by Binet–Cauchy we have \begin{align} \sum_{\mathbf{J}\in\mathcal{F}_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2=(-1)^{m\choose 2}\det AB^t. \end{align} As for $AB^t$, its elements $c_{ij}$ are equal to \begin{align} c_{ij}&=\sum_{k=0}^{m+n-1}a_{i,k}b_{j,k}=(-1)^{m+n-1}\sum_{k=0}^{m+n-1} {-i-1\choose k}{-j-1\choose k^*}\\ &=(-1)^{m+n-1}{-i-j-2\choose m+n-1}= {i+j+m+n\choose m+n-1} \\ &=P((0,-i),(m+n-1,j+1)), \end{align} where, for $u,v\in \mathbb{Z}^2$, $P(u,v)$ is the number of lattice paths from $u$ to $v$ (paths go up and right). So, $\det AB^t$ has Lindström–Gessel–Viennot interpretation. The collection of $m$ non-intersecting lattice paths from the points $u_i:=(0,-i)$ to the points $v_j:=(m+n-1,j+1)$ of course must join $u_i$ with $v_{m-1-i}$. This, in particular, gets $(-1)^{m\choose 2}$ factor out. Also, the path from $v_i$ must start with a horizontal part of length at least $i$; analogously for finishing paths. So, counting these paths is equivalent to counting the non-intersecting paths which go from $(i,-i)$ to $(n+i, m-i)$. These paths correspond to Young diagrams in $m\times n$ boards, and the condition of them being non-intersecting is equivalent to corresponding diagrams be $m$ consecutive sections of a 3d Young diagram in the $m\times m\times n$ box. (This last argument is a standard proof of MacMahon formula).

Since Binet–Cauchy itself has interpretation in terms of Lindström–Gessel–Viennot, you may rephrase this argument more combinatorially, if you prefer. I expect that what is obtained is exactly the answer outlined by Richard Stanley.