Let $u\in C_0^\infty (\overline{\Omega})$, where $\Omega\subset \mathbb{R}^N$ is a bounded domain. Fix some $a\in \Omega$ and choose $r>0$ such that $\overline{\Omega}\subset B(a,r)$.
Define $\phi: B(a,r)\to \mathbb{R}$ by $$\phi(x)=\frac{r^2\|\Delta u\|_\infty}{2N}e^{\frac{|x-a|^2}{r^2}}.$$
Note that $$\Delta \phi(x)=\frac{r^2\|\Delta u\|_\infty}{2N}\left(\frac{4|x-a|^2}{r^4}e^{\frac{|x-a|^2}{r^2}}+\frac{2N}{r^2}e^{\frac{|x-a|^2}{r^2}}\right).$$
As $\Delta\phi(a)=\|\Delta u\|_\infty$ and $\Delta \phi(x)\ge \Delta\phi(a)$, we conclude that $\Delta\phi(x)\ge \|\Delta u\|_\infty$ for $x\in \Omega$. Therefore, $$\Delta (u-\phi)\le \|\Delta u\|_\infty-\|\Delta u\|_\infty=0,\ \forall\ x\in \Omega.\tag{1}$$
On the other hand, $$u(x)-\phi(x)=-\phi(x)\le 0,\ \forall\ x\in \partial \Omega, \tag{2}$$
hence, the maximum principle, combined with $(1)$ and $(2)$ implies that $$u(x)\le \frac{r^2\|\Delta u\|_\infty}{2N}e^{\frac{|x-a|^2}{r^2}},\ \forall x\in \partial \Omega.$$
With an similiar argument, we conclude also that $$ -\frac{r^2\|\Delta u\|_\infty}{2N}e^{\frac{|x-a|^2}{r^2}}\le u(x),\ \forall x\in \partial \Omega,$$
so in the end $$\|u\|_\infty \le\frac{r^2\|\Delta u\|_\infty}{2N} e^t, \forall\ u\in C_0^\infty (\overline{\Omega}).\tag{3} $$
where $t$ depends on $\Omega$ and $e^t>1$. My proof has the constant $e^t$, which depends on $\Omega$, however, I was told that inequality $(3)$, is also true without this constant. Does anyone knows how to prove it?
Remark: $C_0^\infty (\overline{\Omega})$ is the set of all $C^\infty (\overline{\Omega})$, which vanishes on the boundary of $\Omega$.
Remark 2: Note that if $u\in H_0^1(\Omega)$ and $\Delta u\in L^\infty(\Omega)$ then by a density argument, $(3)$ still valid.
The function $$v(x):=u(x) + \frac{\|\Delta u\|_\infty}{2N} |x-a|^2$$ is subharmonic in $\Omega$ and is bounded by $\frac{\|\Delta u\|_\infty}{2N} r^2$ on $\partial\Omega$. Hence $$u\le v\le \frac{\|\Delta u\|_\infty}{2N} r^2$$ Same applies to $-u$.