If $f$ is Lipschitz then $\dim_H f(A) \le \dim_H A$, since $H^s_\delta(f(A)) \le Lip(f) \cdot H^s_\delta(A)$ for any $\delta>0$.

However there exists an absolutely continuous function $f\in W^{1,1}$ such that $\dim_H(f(A)) > \dim_H A$ for some set $A$. Let us construct such a function (by a modification of the classical Cantor function).

Consider the classical one-third Cantor set $T\subset [0,1]$ and one-fifth Cantor set $F \subset [0,1]$. Let $T_n$ and $F_n$ denote respectively the unions of $2^n$ closed segments of lengths $3^{-n}$ and $5^{-n}$ such that $T = \bigcap_{n\in \mathbb N} T_n$ and $F = \bigcap_{n\in\mathbb N} F_n$. The standard computation shows that $\dim_H T = \frac{\ln 2}{\ln 3} > \frac{\ln 2}{\ln 5} = \dim_H F$, so we are going to define $f$ in such a way that $f(F) = T$.

Now let $f_n$ denote the increasing function mapping the endpoints of $F_n$ to the endpoints of $T_n$, interpolated linearly for the rest of $x\in[0,1]$. Arguing as in the construction of the classical Cantor function we get that $f_n$ converges uniformly to a continuous function $f$.
Moreover $f_n'$ converges pointwise on the open set $F^c$ (hence a.e.).

Let $g(x):= \sup_{n\in\mathbb N}|f_n'(x)|$ for a.e. $x$. It is possible to show that
$$
\|g\|_1 \le 1 + \sum_{n=1}^\infty 2^n \cdot 5^{-n} \cdot \frac{3^{-n}}{5^{-n}} < \infty,
$$
hence by dominated convergence $f'\in L^1[0,1]$, i.e. $f$ is absolutely continuous (and $f\in W^{1,1}$). Since $f(F_n) = T_n$ for all $n$ it follows that $f(F) = f(T)$.