**Summary**

I don't think that this leads to a generalization of determinant. On the other hand, I see a clear connection to *permanent* (which, if I understand correctly, yields the same tropicalization). That is because the OP's formula can be rewritten as an immanant-like expression evaluated on an auxiliary matrix $\widehat{A}$ obtained by repeating the rows of $A$ according to the multiplicities $m_i.$

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**Analysis**

Up to a constant multiple, the procedure outlined in the question amounts to the following.

>1. Replace the rectangular $m\times n$ matrix $A$ with the square $n\times n$ matrix $\widehat{A}$ by repeating the $i$th row of $A$ $m_i$ times.

>2. Consider the row expansion of the "immanant" 
$\sum_{\sigma\in S_n}k_{\sigma}\prod_{i=1}^n \widehat{A}_{i,\sigma(i)},$ where $k_\sigma$ are some coefficients that need to be specified ("the signs").

The partition $\{I_1,I_2,\ldots,I_m\}$ from the original formula is $\large\{\sigma$(first $m_1$ indices), $\sigma$(next $m_2$ indices), $\ldots$, $\sigma$(last $m_m$ indices)$\large\}$ and $\operatorname{sgn}\mathcal{I}=\sum k_\sigma$ over all $\sigma$ corresponding to that $\mathcal{I}.$ 

In the case $k_\sigma=1$ for all $\sigma\in S_n,$ we get the permanent of $\widehat{A}$ divided by $\prod_{i=1}^{m}m_i!$

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**Examples.**

1. For $m=1$, the matrix $A$ is a single row, $A=[a_1,a_2,\ldots,a_n],$ the multiplicity $m_1=n,$ there is only one possible partition, where $I_1=\{1,2,\ldots,n\},$ and (assuming the coefficient is $1$) the OP's expression is $\prod_{i=1}^n a_i.$ The matrix $\widehat{A}$ is the row $[a_1,a_2,\ldots,a_n]$ repeated $n$ times and its permanent is $n!$ times the product of the entries of $A.$ Unless $n=1,$ the determinant of $\widehat{A}$ is $0$ (corresponding to the "sign" coefficient $0$) and it's hard to argue that the product of the entries of $A$ should be an analogue of the determinant.

2. For $m=2,$ the matrix $A$ has two rows and the multiplicity is a pair $(p,q)$ of natural numbers that add to $n.$ The OP's procedure is to choose $p$ entries in the first row of $A,$ which determine $q$ elements in the second row so that every column is used exactly once, multiply the entries together, and take a linear combination of these products over different choices with unspecified coefficients. It is easy to see that this is essentially the same as repeating the first row of $A$ $p$ times and the second row $q$ times to form a square matrix $\widehat{A}$ and computing its "immanant".

3. For $m=n,$ the multiplicities are necessarily all equal to $1$ and partition $\mathcal{I}$ is the same as a permutation $\sigma$ of $\{1,2,\ldots,n\}.$ In this case, the recipe amounts to the usual row expansion of the "immanant" of $A$, which gives the determinant or the permanant if the coefficients $k_\sigma$ are $\operatorname{sgn}\sigma$ or $1,$ respectively. Of course, in this case there are also other possibilities.