Does a monotone convergence theorem for $L^\infty$ norm holds?

YES.

First of all, since $f_n$ is non-negative and non-decreasing, so is $|f_n|$. Hence $$ |f_1|_\infty\le |f_2|_\infty\le \cdots\le |f_n|_\infty \le \sup_{n\in\mathbb N}|f_n|_\infty=\lim_{n\to\infty}|f_n|_\infty $$ Meanwhile $$ f_n(x)\le f(x),\quad\text{for all $n\in\mathbb N$ and $n\in\mathbb N$}, $$ and hence $|f_n|_\infty\le|f|_\infty$, and consequently $$ \lim_{n\to\infty}|f_n|_\infty\le |f|_\infty\tag{1} $$ If $\lim_{n\to\infty}|f_n|_\infty=\infty$, then clearly $\lim_{n\to\infty}|f_n|_\infty= |f|_\infty$.

Assume now that $\lim_{n\to\infty}|f_n|_\infty=M<\infty$.

As $|f_n|_\infty$ is non-decreasing, this means that for every $n\in\mathbb N$, $$ 0\le f_n(x)\le M, \quad\text{for all $x\in X$}. $$ Hence $$ 0\le f(x)=\lim_{n\to\infty}f_n(x)\le M, \quad\text{for all $x\in X$}. $$ and finally $$ |f|_\infty\le M=\lim_{n\to\infty}|f_n|_\infty. \tag{2} $$ Now combine $(1)$ and $(2)$.

The answer is YES. Clearly, $f_n$ increasing to $f$ implies $\|f_n\|_{\infty} \leq \|f\|_{\infty}$. To prove the other way note that $|f(x)| \leq \sup |f_n(x)| \leq \sup \|f_n\|_{\infty}$ almost everywhere. Hence $\|f\|_{\infty} \leq \sup \|f_n\|_{\infty}=\lim \|f_n\|_{\infty}$.