First, we should show that for $A\in \mathbb{R}^{m\times n}$: $\|Av\|_1 \le \|A\|_1 \|v\|_1$.
$$(1)\quad\|Av||_1=\sum_{i=1}^m|(Av)_i| \le \sum_{i=1}^m \sum_{j=1}^n|A_{i j}||v_j|=\sum_{j=1}^n (\sum_{i=1}^m |A_{ij}|)|v_j|\le \|A\|_1 \|v\|_1$$
Is that even correct? Because next, we are supposed to construct a vector such that $\|Av\|_1 = \|A\|_1 \|v\|_1$ for some fixed $A$.
Constructing a vector such that the first inequality in (1) becomes equal is simple (equal triangle inequality).
But the second inequality in (1) has me stumped. We can't possibly guarantee that all column sums of $A$ are equal, so how can I find that vector?
From the discussion in comments, you define $$\| A \|_1 = \max\{\|c(A,j)\|_1 \mid j=1,\ldots,n\},$$ where $c(A,1),\ldots,c(A,n)$ are the columns of $A$. Now, let $k$ be such that $\|c(A,k)\|_1=\|A\|_1$ and let $e_k$ be the $k$-th vector of the canonical basis of $\Bbb R^n$, i.e. $(e_k)_i=0$ if $i\neq k$ and $(e_k)_k = 1$. Then, we have $\|e_k\|_1=1$ and $$\|Ae_k\|_1 = \|c(A,k)\|_1=\|A\|_1=\|A\|_1\|e_k\|_1.$$
Side note: In conclusion you have then shown that $$\|A\|_1=\max_{x\neq 0}\frac{\|Ax\|_1}{\|x\|_1}.$$