There is at least an inequality between the product of all entries of the matrix and the permanent of a matrix with non-negative entries. The geometric-arithmetic inequality states that $(x_1\dots x_n)^{1/n}\leq\frac{1}{n}(x_1+\dots+x_n)$ whenever $x_1,\dots,x_n$ are non-negative. Therefore, $n(x_1\dots x_n)^{1/n}\leq x_1+\dots+x_n$ whenever $x_1,\dots,x_n$ are non-negative.
Suppose $A$ is a matrix with non-negative entries. Then
$$\text{per}(A)=\sum_{\sigma\in S_{n}}\prod_{k=1}^{n}a_{k,\sigma(k)}\geq n!\cdot(\prod_{\sigma\in S_{n}}\prod_{k=1}^{n}a_{k,\sigma(k)})^{1/n!}=n!\cdot\big((\prod_{i,j}a_{i,j})^{(n-1)!}\big)^{1/n!}$$ $$=n!\cdot(\prod_{i,j}a_{i,j})^{1/n}.$$
Here, equality is reached if and only if every entry in $A$ is the same.