$A$ has the same eigenvalues as its transpose, that I will denote $B$. For $B$, the hypothesis means that $\sum_{j=1}^n|a_{ij}|\leq 1$ for all $i$. If $x\neq 0$ is an eigenvector for $\lambda$, and $i$ is such that $x_i=\lVert x\rVert_{\infty}$, then we have $\sum_{j=1}^na_{ij}x_j=\lambda x_i$ hence $\sum_{j\neq i}a_{ij}x_j=(\lambda-a_{ii})x_i$ and $\lVert x\rVert_{\infty}|\lambda-a_{ii}|\leq \sum_{j\neq i}|a_{ij}|\cdot |x_j|\leq \sum_{j\neq i}|a_{ij}|\cdot\lVert x\rVert_{\infty}\leq (1-|a_{ii}|)\cdot\lVert x\rVert_{\infty}.$ As $x\neq 0$, we get $|\lambda-a_{ii}|\leq 1-|a_{ii}|$ hence $|\lambda|\leq 1$.
(we proof the Gershgorin circle theorem)