**Summary**

There is a clear connection to the [permanent](https://en.wikipedia.org/wiki/Permanent), which, if I understand correctly, yields the same tropicalization, and a more speculative connection to the determinant. The reason is that OP's formula can be rewritten as an [immanant](https://en.wikipedia.org/wiki/Immanant_of_a_matrix)-like expression evaluated on an auxiliary matrix $\widehat{A}$ obtained by repeating the rows of $A$ according to the multiplicities $m_i.$ For arguably the most natural choice of the coefficients in the expansion, where they are all equal $1$, one obtains the *permanent* of $\widehat{A}$ (up to a constant multiple). On the other hand,  one can also choose the coefficients for an arbitrary $m\leq n$ in a way that reproduces the determinant when $m=n.$ I don't know whether this leads to a meaningful generalization of the determinant. 

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**Analysis**

Modulo the choice of the coefficients, 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 $\mathcal{I}=\{I_1,I_2,\ldots,I_m\}$ of the set $\{1,2,\ldots,n\}$ of the column indices from the original formula is $\large\{\sigma$(first $m_1$ indices), $\sigma$(next $m_2$ indices), $\ldots$, $\sigma$(last $m_m$ indices)$\large\}$ and the coefficient $\operatorname{sgn}(\mathcal{I})=\sum k_\sigma,$ with the sum taken over all permutations $\sigma$ corresponding to that partition $\mathcal{I}.$ 

In the case that $k_\sigma=1$ for all $\sigma\in S_n,$ we get the permanent of $\widehat{A},$ which is equal to the constant $(\prod_{i=1}^{m}m_i!)$ times (OP's expression for all coefficients equal to $1$). Although this is non-standard in the context of permanents, the constant can be eliminated by setting $k_{\sigma}=0$ for permutations $\sigma$ that "scramble" some of the parts $I_k$, cf the next paragraph.

One can also consider the following "determinant" choice of $k_\sigma.$ It is defined to be $\operatorname{sgn}(\sigma)$ if the descent set of $\sigma$ is contained in $\{m_1,m_1+m_2,\ldots,m_1+\ldots+m_{m-1}\}$ and $0$ otherwise. In other words, for each partition $\mathcal{I}$ consider the element $\sigma_{\mathcal{I}}\in S_n$ which maps the first $m_1$ indices to $I_1$ *preserving the order*, the next $m_2$ indices to $I_2$ *preserving the order*, and so on, and set $\operatorname{sgn}(\mathcal{I})=\operatorname{sgn}(\sigma_{\mathcal{I}}).$ The resulting expression is always non-trivial and for $m=n,$ it yields $\det(A).$ On the other hand, if $m<n$ then the matrix $\widehat{A}$ contains repeating rows and hence its determinant is $0.$

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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 $\prod_{i=1}^n a_i,$ that is the product of the entries of $A.$ 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 $A.$ Even though the "determinant" choice of the "sign" coefficient is $1,$ it appears more natural to view the product of the entries of $A$ as an analogue of the permanent rather than the determinant. On the other hand, unless $n=1,$ the determinant of $\widehat{A}$ is $0$ (corresponding to the "sign" coefficient being $0$).

2. For $m=2,$ the matrix $A$ has two rows and $n$ columns and the multiplicity is a pair $(p,q)$ of natural numbers that add up to $n.$ A partition $\mathcal{I}=\{I,J\}$ of the set of column indices with $|I|=p$ and $|J|=q$ leads to the choice of $p$ entries in the first row of $A$ and $q$ entries in the second row in such a way that every column contains exactly one chosen entry. OP's procedure is to multiply the chosen entries together and take a linear combination of these products over different choices with to-be-specified coefficients $\operatorname{sgn}(I,J).$ It is easy to see that this is effectively 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" as above. There are two natural choices for the coefficients: $\operatorname{sgn}(I,J)=1$, leading to the permanent of $\widehat{A},$ and $\operatorname{sgn}(I,J)=\operatorname{sgn}(\sigma),$ where $\sigma$ is the *shortest* permutation corresponding to the partition $(I,J)$, namely $\sigma(i)=I_i$ for $1\leq i\leq p$ and $\sigma(p+i)=J_{i}$ for $1\leq i\leq q$ (it is assumed that the parts $I$ and $J$ are represented by *increasing* sequences of indices).

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 permanent if the coefficients $k_\sigma$ are $\operatorname{sgn}\sigma$ or $1,$ respectively. Of course, there are also other possibilities, for example, $k_{\sigma}=\chi_{\lambda}(\sigma),$ where $\chi_{\lambda}$ is an irreducible character of the symmetric group $S_n,$ corresponds to the immanant $\operatorname{Imm}_{\lambda}(A)$ of $A.$