This seems too strong, regardless of which set $\omega \subset \Omega$ one works with. We suppose that $\omega$ has bounded perimeter, so that $\chi_\omega \in BV(\Omega)$.

Let $(f_n \mid n \in \mathbf{N})$ be a sequence of functions in $W^{1,1}(\Omega)$ with
\begin{equation}
\lvert f_n \rvert_{L^1} \to 0
\text{ and }
V(f_n) \geq 3 \, \mathrm{Per}(\omega).
\end{equation}
Let moreover $g_n \in W^{1,1}(\Omega)$ be a sequence of functions
so that 
\begin{equation}
\lvert g_n - \chi_\omega \rvert_{BV} \to 0;
\end{equation}
for example $g_n$ can be obtained by a mollification argument. Then the two sequences combined have
\begin{equation}
\lvert f_n + g_n - \chi_\omega \rvert_{L^1} \to 0 
\text{ but }
V(f_n + g_n) \geq V(f_n) - V(g_n) \geq 3/2 \mathrm{Per}(\omega).
\end{equation}


We construct Lipschitz functions $f_n$ on $\Omega$ that are independent of $y$. Specifically let, given $A > 0$,
\begin{equation}
f_n(x,y) =\begin{cases}
Ax & \text{ on $[0,\frac{1}{2n}]$} \\
A/n - Ax & \text{ on $[\frac{1}{2n},\frac{1}{n}]$}
\end{cases}
\end{equation}
and extend this to be $\frac{1}{n}$-periodic in $x$.

Then $\lvert f_n \rvert_{L^\infty} \leq \frac{A}{2n} \to 0$ but the total variation is constant equal to $\int_\Omega \lvert D f_n \rvert = A \lvert \Omega \rvert$. To conclude it only remains to take $A > 3 \mathrm{Per}(\omega)/\lvert \Omega \rvert$.