As t.b. noted, we are looking at the functions $f_p:{\mathbb R}^n\to {\mathbb R}$ given by $f_p(x)=\|x\|_p$ and want to know if they converge to $f_\infty$ in some norm on $C({\mathbb R}^n,{\mathbb R})$. It is easy to see that $f_p \to f_\infty$ uniformly on compact sets, and so if you consider the space $C(\Omega,{\mathbb R})$ where $\Omega \subset {\mathbb R}^n$ is bounded and open, then $f_p \to f_\infty$ in, for example, all $L^p$ norms on $C(\Omega,{\mathbb R})$.
For $C({\mathbb R}^n,{\mathbb R})$, the problem is that the the functions $f_p$ do not belong to $C({\mathbb R}^n,{\mathbb R})$, when you endow it with any of the standard norms, such as $L^p$ norms, so we cannot even talk about convergence to $f_\infty$. That being said, you can look at less common norms such as
$\|f\|_X = \sup_{r >0}\left( e^{-r}\sup_{|x|\leq r} |f(x)|\right)$
or weighted $L^P$ norms
$\displaystyle \|f\|_{w,p} = \left( \int_{{\mathbb R}^n} w(x) |f(x)|^p dx\right)^{1/p}$
where $w$ is positive weighting function which decays to zero (exponentially) as $|x|\to \infty$ in order to "cancel out" the growth of $f$. Then the sequence $f_p$ would actually belong to the space $C({\mathbb R}^n,{\mathbb R})$ when endowed with either of these norms. I haven't written this down, but it looks like it would be very easy to show that $\|\cdot\|_p \to \|\cdot\|_\infty$ in both $\|\cdot\|_X$ and $\|\cdot\|_{w,p}$
As you already noted, $(\|u\|_2+\|\Delta u\|_2)^{1/2}$ is an equivalent norm on $H^2$, hence there is a constant $c>0$ such that
$$c\|u\|_{H^2}^2 \leq \|u\|_2^2+\|\Delta u\|_2^2, \quad \forall u \in H^2.$$
This is also true for $u \in \tilde V \subset H^2$. Since the elements in $\tilde V$ have zero mean, we can make use of the following Poincare inequality,
$$\|u\|_2^2 \leq C_P\|\nabla u\|^2, \quad \forall u \in \tilde V. $$
Putting these two inequalities together, we have for all $u \in \tilde V$,
$$c\|u\|_{H^2}^2 \leq \|u\|_2^2+\|\Delta u\|_2^2 \leq C_P\|\nabla u\|_2^2 + \|\Delta u\|^2_2 \leq C (\|\nabla u\|_2^2 + \|\Delta u\|^2_2) \leq C \|u\|_{H^2}^2,$$
where $C:=\max\{1,C_P\}$. Hence, $(\|\nabla u\|_2^2 + \|\Delta u\|_2^2)^{1/2}$ is an equivalent norm on $\tilde V$.
Best Answer
$$\int_{\Omega}|f|^p \le \int_{\Omega}||f||_{\infty}^p = m(\Omega) ||f||_{\infty}^p$$ hence, taking $p$-th roots, $$||f||_p \le m(\Omega)^{1/p} ||f||_{\infty}$$
Finally, $c=m(\Omega)^{1/p}$ is optimal, since the inequality becomes an equality for constant function $f$.