If the $f_n$ are absolutely continuous, then $f'_n$ would also be the distributional derivative and $g=f'$ follows from the continuity of that. So if there is a counterexample it should probably involve the Cantor-parts of a sequence of BV function converging to something absolutely continuous.

Indeed I think the following should work: Let $f_0:\mathbb{R} \to \mathbb{R}$ be the standard Cantor-staircase, repeated so that $f(x+1)=f(x)+1$. Define $f_k(x) := \frac{1}{k}f_0(kx)$. Then $f_k$ converges to $f(x)=x$ uniformly. But $f_k'(x) =0$ a.e. for all $k$, so $g(x) = 0 \neq 1 = f'(x)$ almost everywhere.

If the $f_n$ are absolutely continuous, then $f'_n$ would also be the distributional derivative and $g=f'$ follows from the continuity of that. So if there is a counterexample it should probably involve the Cantor-parts of a sequence of BV function converging to something absolutely continuous.

Indeed I think the following should work: Let $f_0:\mathbb{R} \to \mathbb{R}$ be the standard Cantor-staircase, repeated so that $f_0(x+1)=f_0(x)+1$. Define $f_k(x) := \frac{1}{k}f_0(kx)$. Then $f_k$ converges to $f(x)=x$ uniformly. But $f_k'(x) =0$ a.e. for all $k$, so $g(x) = 0 \neq 1 = f'(x)$ almost everywhere.