Yes, we can get $\exp(\omicron(n))$.
Assume for a moment that $n$ is a perfect square and $m=\sqrt{n}$. The general case is essentially the same, just a bit more complicated. The idea is to partition those $n=m^2$ variables $y_1,\ldots\, y_n$ in $m$ blocks of $m$ digits in base $b=n+1$. Define
$$a_{ij} = \left\{ \begin{array}{ll} b^{\,j-m(i-1)-1} & \text{if } \; m(i-1) \lt j \leq mi \\ 0 & \text{otherwise} \end{array} \right. $$
For example if $n=9$ we have $m=3$ and $b=10$ and
$$ A = \left( \begin{array}{rrr|rrr|rrr} 1 & 10 & 100 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 10 & 100 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 10 & 100 \\ \end{array} \right) $$
It is not hard to see that such $A$ only admits the trivial solution where all the $y_j$ are equal to $1$, because $0 \le y_j \lt n + 1 = b$.
Now $\max a_{ij} = b^{m-1} \le b^m$ and thus
$$ 2^m \cdot \max a_{ij} \le 2^m b^m = (2b)^m = \exp(m \, \ln(2b)) = \exp(\sqrt{n} \,\ln(2n+2)) $$
and $\sqrt{n} \,\ln(2n+2)$ is in $\omicron(n)$.
The question actually asked for positive integer entries $a_{ij}$ rather than just non-negative entries. But because $\sum_{j=1}^n y_j = n$, we can instead use $a'_{ij} = a_{ij} + 1$, and since $\,\max a'_{ij} = b^{m-1} + 1 \le b^m$ we arrive at the same bound.
Edit (based @joro's comment) In summary, we have $\sqrt{n}$ equations with $\max a_{ij} \sim n^{\sqrt{n}}$.