Number of integers covered Let $x_{ij}\in\{0,1\}$ for ${i=1}$ to ${m}$ and for ${j=1}$ to $n$. How many different values does $$\prod_{i=1}^m\sum_{j=1}^nx_{ij}$$ cover?

Is there an $a_{ij}$ such that $$\prod_{i=1}^m\sum_{j=1}^na_{ij}x_{ij}$$ for which we cover least number of distinct values as compared to that values covered by first formula?
For $m=2$ this is one more (for the $0$) than the number of distinct products $s \cdot t$ with $s,t \in \{1,2,\ldots,n\}.$ The latter number is for $n=1,2,...$ the sequence $1,3,6,9,14,18$ which certainly no simple formula is known to produce. It is discussed as an o.e.i.s. sequence A027424.
For larger $m$ it would seem unlikely for there to be a closed formula, since even for $m=2$ there doesn't seem to be a known one. However note one of the links from the o.e.i.s. is to a paper discussing the more general case of an arbitrary number of factors. (Unfortunately that link only lets one see an outline, and refers to yet another site where the final paper might be found.)
Added note about the second part of the question about $a_{ij}$ choice: It may not be optimal, but choosing $a_{ij}=(-1)^{j+1}$ certainly brings the count down. Suppose $n=2k$ is even. Then the sums for a fixed row $j$ of the terms $a_{ij}x_{ij}=(-1)^{j+1}x_{ij}$ have as possible values the integers in the interval $[-k,k].$ If we omit the zero choice here (later one can add $1$ to the sum) then the possible values of the product are each up to a sign of $\pm 1$ the nonzero values of the original unaltered sum for parameters $(m,n)=(m,k).$ To state this more clearly, let $1+S(m,n)$ be the number covered as in the post, while $1+T(m,n)$ the number covered with our choice of coefficients $a_{ij}.$ Then what we have is $T(m,2k)=2 \cdot S(m,k).$ Even for $m=2$ this gives a significant decrease for the altered sum, comared to the unaltered one.
I suspect this is difficult.  But here is an upper bound.
Each expression is the product of $m$ numbers, each between $0$ and $n$.  If any number is zero, the product is zero.  Otherwise, there are $m$ numbers, each between $1$ and $n$.
By the standard 'stars and bars' argument, there are $m+n-1\choose n-1$ sets, so an upper bound is $$1+{m+n-1\choose n-1}$$
Of course, that doesn't take into account coincidences such as $1*4=2*2$.
A lower bound comes with all $a_i$ primes greater than $\sqrt{n}(i=1..m)$.  All these products are different.  Let $q=\pi(n)-\pi(\sqrt{n})\approx n/\log n$.  The lower bound is $$1+{m+q-1\choose m}$$