The inequality is obviously false, because if $f\equiv c$ is a constant, then $u\equiv c$ remains constant. Since $\Omega$ is bounded, this $f$ is $L^2(\Omega)$, and yet $\|u(t)\|_\infty=|c|$ does not decay as $t\to+\infty$.

I suppose that you make a confusion with the heat equation in the hole space ($\Omega={\mathbb R}^N$, of course the boundary condition disappears). Then
$$u(t)=H_t\star f$$
where the heat kernel is given by
$$H_t(x)=\frac1{(2\pi t)^{N/2}}K\left(\frac x{\sqrt t}\right),\qquad K(y):=\exp\frac{-|y|^2}4.$$
From Young inequality (or just Hölder)
$$\|u(t)\|_\infty\le\|H_t\|_2\|f\|_2=\frac c{T^{N/4}}\|f\|_2.$$

Turning back to the case of a bounded domain, the general rule is an exponential decay as $t\to+\infty$. For instance, the spectrum of $\Delta$ under the Dirichlet boundary condition $u|_{\partial\Omega}=0$ is real, bounded above by $-\mu_D<0$, and we get $\|u(t)\|_p=O(e^{-\mu_D t})$ for every $p\ge2$. In the case of the Neumann boundary condition, we must subtract the mean of the initial data:
$$\|u(t)-\frac1{|\Omega|}\int_\Omega f(x)dx\|_p=O(e^{-\mu_N t}),\qquad\forall p\ge2.$$
Here $\mu_N$ is the least non-zero eigenvalue of $-\Delta$.

depending on $T$. This form of ultracontractivity is true for the Neumann Laplacian on bounded domains with Lipschitz boundary. $\endgroup$4more comments