A typical counterexample is like the following. For concreteness, let's take $p=1$ and $n=2$, and $\Omega = B(0,1) \subset \mathbb{R}^2$.
Let $$f_n(t) = \begin{cases} nt, & 0 \le t \le 1/n \\ 1, & t > 1/n.\end{cases}$$
Set $v_n(x) = f_n(|x|)$, so $v_n$ is continuous and $v_n(0)=0$. (If you like you may modify the function $f_n$ slightly to make it $C^\infty$.)
You can check that $|\nabla v_n(x)| = n$ for $0 < |x| < 1/n$ and $|\nabla v_n(x)| = 0$ for $1/n < x < 1$. Thus $|v_n|_{W^{1,1}} = n \cdot m(B(0,1/n)) = \pi/n \to 0$. But $v_n \to 1$ monotonically almost everywhere, so $v_n \to 1$ in $L^1(\Omega)$.
The flaw in your proof is in looking at the properties of $v$. You defined $v$ as the $L^p$ and a.e. limit of the $v_n$, so it's only well-defined up to null sets. In general you can't choose a continuous representative for it (you claimed you could but didn't justify it), so speaking of $v(0)$ doesn't really make sense. In this example you can choose a continuous representative (namely 1) but you don't have $v_n \to v$ pointwise everywhere, so using that representative you cannot conclude $v(0)=0$.
$\newcommand\Om\Omega$Now consider the general case of any natural $d$. Here we will give an upper bound on $\|f\|_\infty$ in terms of $\|f\|_1$, $L$, and $d$. This bound will be optimal up to a factor depending only on $d$; as follows from a comment of yours, such factors do not matter to you. The mentioned bound will be exact in the case $d=1$.
Indeed, let $I:=[0,1]$, $\Om:=I^d$,
$$M:=\|f\|_\infty=\max_{x\in\Om}|f(x)|=|f(a)|$$
for some $a=(a_1,\dots,a_d)\in\Om$. Then
\begin{equation}
|f(x)|\ge h_a(x):=(M-L|x-a|)_+=L(r-|x-a|)_+ \tag{1}
\end{equation}
for all $x\in\Om$, where $|x-a|$ is the Euclidean norm of $x-a$, $u_+:=\max(0,u)$, and
$$r:=M/L$$
(assuming $L>0$). So,
$$\frac{\|f\|_1}L=\frac1L\int_\Om|f|\ge\int_\Om dx\,(r-|x-a|)_+
=E(r-R)_+,$$
where
$$R:=\sqrt{\sum_1^d(U_i-a_i)^2}$$
and $U_1,\dots,U_d$ are independent random variables each uniformly distributed on $[0,1]$.
Next,
$$E(r-R)_+=E\int_0^r dv\,1(R<v)=\int_0^r dv\,P(R<v),$$
\begin{align*}
P(R<v)&=P\Big(\sum_1^d(U_i-a_i)^2<v^2\Big) \\
&\ge\prod_1^d P(|U_i-a_i|<v/\sqrt d) \tag{2} \\
&\ge\prod_1^d P(|U_i|<v/\sqrt d) \tag{3} \\
&=\min\Big(1,\frac{v}{\sqrt d}\Big)^d=:Q(v).
\end{align*}
So,
\begin{align*}
\frac{\|f\|_1}L&\ge\int_0^r dv\,P(R<v) \\
&\ge\int_0^r dv\,Q(v) \\
&=g(r):=\left\{\begin{aligned}
\frac{r^{d+1}}{c_d^{d+1}}&\text{ if }r\le\sqrt d, \\
r-\frac{d\sqrt d}{d+1}&\text{ if }r\ge\sqrt d,
\end{aligned}
\right.
\end{align*}
where
\begin{equation*}
c_d:=((d+1)d^{d/2})^{1/(d+1)}.
\end{equation*}
Solving now the inequality $\frac{\|f\|_1}L\ge g(r)$ for $r$ and
recalling that $r=M/L=\|f\|_\infty/L$, we get
\begin{align*}
\|f\|_\infty&\le B_0(\|f\|_1,L) \\
&:=Lg^{-1}\Big(\frac{\|f\|_1}L\Big) \\
&=\left\{\begin{aligned}
c_d L^{d/(d+1)}\|f\|_1^{1/(d+1)}&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\
%\|f\|_1&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\
\|f\|_1+\frac{d\sqrt d}{d+1}\,L&\text{ if }\|f\|_1\ge\frac{\sqrt d}{d+1}\,L.
\end{aligned}
\right.
\end{align*}
Remark 1: Obviously, the inequality in (1) will turn into the equality if we choose $f=h_a$ with $a=0$, and then the inequality in (3) will turn into the equality as well. Moreover, the inequality in (2) will change the direction if we replace $v/\sqrt d$ in (2) by $v$.
Therefore, the bound $B_0(\|f\|_1,L)$ is optimal up to a factor depending only on $d$.
It also follows that the bound $B_0(\|f\|_1,L)$ is exact when $d=1$, in which case $B_0(\|f\|_1,L)$ is
exactly the same as the exact upper bound on $\|f\|_\infty$ presented in the other answer of mine on this web page (previously obtained somewhat differently).
Remark 2: We have $\|f\|_1\le\|f\|_2$, since the Lebesgue measure on $\Om$ is a probability measure. Also, $B_0(\cdot,L)$ is nondecreasing. So,
$$\|f\|_\infty\le B_0(\|f\|_1,L)\le B_0(\|f\|_2,L).$$
Remark 3: One can show that $c_d\le\sqrt{2d}$ for all natural $d$.
Remark 4:
It follows from Remark 3 that
\begin{align*}
\|f\|_\infty&\le B_1(\|f\|_1,L) \\
&:=\left\{\begin{aligned}
\sqrt{2d}\, L^{d/(d+1)}\|f\|_1^{1/(d+1)}&\text{ if }\|f\|_1<\frac{\sqrt d}{d+1}\,L, \\
%\|f\|_1&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\
(d+1)\|f\|_1&\text{ if }\|f\|_1\ge\frac{\sqrt d}{d+1}\,L.
\end{aligned}
\right.
\end{align*}
Note that $B_1(\|f\|_1,L)$ differs from $B_0(\|f\|_1,L)$ by, at most, a factor depending only on $d$. So, in view of Remark 1, the bound $B_1(\|f\|_1,L)$ is optimal as well up to a factor depending only on $d$.
Note also that the exponent of $\|f\|_1$ in the bound $B_1(\|f\|_1,L)$ is $1/(d+1)$ if $\|f\|_1$ is not too large as compared with $L$, and this exponent is $1$ otherwise.
In view of Remark 2, we also have
$$\|f\|_\infty\le B_1(\|f\|_2,L).$$
Remark 5: As shown in Willie Wong's comment, if we had a bound on $\|f\|_\infty$ of the form $C(d)L^a\|f\|_1^b$, then the only possible values for $a$ and $b$ would be $d/(d+1)$ and $1/(d+1)$, respectively. However, in view of Remark 4, it is clear that such a bound on $\|f\|_\infty$ is impossible: the exponent of $\|f\|_1$ cannot be greater than $1/(d+1)$ for values of $\|f\|_1$ not too large as compared with $L$, and this exponent cannot be less than $1$ for values of $\|f\|_1$ large enough as compared with $L$.
Best Answer
For $p>\max(2,d)$ you have by Sobolev embedding $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|f-\overline{f}\|_p + \|\nabla f\|_p.$$ The interpolation $L^p(\Omega) = [L^2(\Omega),L^\infty(\Omega)]_\theta$ with $\theta\in(0,1)$ entails $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|f-\overline{f}\|_2^\theta \|f-\overline{f}\|_\infty^{1-\theta} + \|\nabla f\|_2^\theta\|\nabla f\|_\infty^{1-\theta}.$$ Now, you can absorb part of the first term of r.h.s. in the l.h.s. with Young's inequality $a b \leq \theta a^{1/\theta} + (1-\theta) b^{1/(1-\theta)}$ : $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|f-\overline{f}\|_2 + \|\nabla f\|_2^\theta \|\nabla f\|_\infty^{1-\theta}.$$ The usual Poincaré-Wirtinger inequality allows to write $$\|f-\overline{f}\|_\infty \lesssim_{\Omega,p} \|\nabla f\|_2 + \|\nabla f\|_2^\theta \|\nabla f\|_\infty^{1-\theta},$$ and this implies what you hope for, because you can write (but this is rough ...) $$\|\nabla f\|_2 = \|\nabla f\|_2^\theta \|\nabla f\|_2^{1-\theta}\leq \|\nabla f\|_2^\theta \|\nabla f\|_\infty^{1-\theta}.$$