The proof is essentially same as here. The matrix does not need to be triangular; only a row-wise diagonal dominance is needed.
Assume $U$ is a square invertible matrix such that
$$\tag{1}
|u_{ii}|-\alpha\sum_{j\neq i}|u_{ij}|\geq 0, \quad i=1,\ldots,N,
$$
for some $\alpha>1$. Let $x$ be an $N$-vector and $i$ be the index such that $\|x\|_\infty=|x_i|$. Then
$$
\begin{align}
\|Ux\|_\infty\geq|(Ux)_i|&\geq|u_{ii}||x_i|-\sum_{j\neq i}|u_{ij}||x_j| &&\text{(triangle inequality)} \\
&\geq|u_{ii}||x_i|-\sum_{j\neq i}|u_{ij}||x_i| && \text{(since $|x_j|\leq|x_i|=\|x\|_\infty$)}\\
&\geq\left(1-\frac{1}{\alpha}\right)|u_{ii}||x_i|&&\text{(using (1))}
\\&\geq\left(1-\frac{1}{\alpha}\right)\min_i|u_{ii}|\|x\|_\infty&&\text{(making bound independent of $i$).}
\end{align}
$$
Hence
$$
\frac{\|Ux\|_\infty}{\|x\|_\infty}\geq
\frac{\alpha-1}{\alpha}\min_i|u_{ii}|.
$$
We have then
$$\tag{2}
\|U^{-1}\|_\infty=\max_{x\neq 0}\frac{\|U^{-1}x\|_\infty}{\|x\|_\infty}=\max_{x\neq 0}\frac{\|x\|_\infty}{\|Ux\|_\infty}\leq\frac{\alpha}{\alpha-1}\frac{1}{\min\limits_i|u_{ii}|}.
$$
This already gives the bound you seek with $\alpha=2$.
Note that (1) also implies a bound on $\|U\|_\infty$ in terms of the diagonal entries $u_{ii}$ and hence a more neat bound on $\kappa_{\infty}$ :-). We have
$$
|u_{ii}|+\sum_{j\neq i}|u_{ij}|\leq\left(1+\frac{1}{\alpha}\right)|u_{ii}|, \quad i=1,\ldots,N
\quad\Rightarrow\quad
\|U\|_{\infty}\leq\frac{\alpha+1}{\alpha}\max_i|u_{ii}|,
$$
so with (2), we get
$$
\kappa_\infty(U)\leq\frac{\alpha+1}{\alpha-1}\frac{\max\limits_i|u_{ii}|}{\min\limits_i|u_{ii}|}.
$$
