Bounding the 1-norm of a matrix Let $M$ be a real $n\times n$ matrix and let $\|M\|_1 = \sum_{ij} |M_{ij}|$ ("the entry-wise 1-norm").
My question is: How well can we bound $\|M\|_1 $ in terms of $|\det M|$ (from below)?
Here is an obvious bound: Let $m = \operatorname{max}_{ij} |M_{ij}|$. 
Then it is obvious that $$|\det M| \leq \sum_{\sigma \in S_n} m^n = n! m^n \quad  or \quad  \|M\|_1  \geq m \geq (|\det M|/n!)^{\frac1n}.$$ Can this bound be improved?
(In fact, the above result is strong enough for my purposes, but this question came out of curiosity. I have played around for a while but I couldn't improve the bound, nor achieve it.)

If we denote by $v_1,\ldots,v_n$ the column of $M$, we have $|\det M|\leq \prod_{j=1}^n\lVert v_j\rVert_2$. But $\lVert v_j\rVert_2^2 =\left(\sum_{k=1}^n|m_{kj}|\right)^2-2\sum_{j<k}|m_{jk}|\leq \left(\sum_{k=1}^n|m_{kj}|\right)^2$ hence $\lVert v_j\rVert_2\leq \sum_{k=1}^n|m_{jk}|\leq \lVert M\rVert_1$. We finally get that $|\det M|^{\frac 1n}\leq \lVert M\rVert_1$ which is a better bound. 
