The classical Poincaré inequality states that
Let $1 \le p < \infty$ and $U$ be an open bounded subset of $\mathbb{R}^n$. Then there exists a constant $C$ (depending only on $U$ and $p$) such that for any function $u\in W^{1,p}_0(U)$, $$\|u\|_{L^{p}(U)}\leq C\|\nabla u\|_{L^{p}(U)}$$
Let $\Omega=\prod_{i=1}^n[a_i, b_i]$ such that $U\subseteq\Omega\subset\mathbb{R}^n$. Then \begin{align} \int_U |u(x)|^pdx &\stackrel{(1)}{=} \int_\Omega |u(x)|^pdx \\ &= \int_\Omega \left|\int_{t=a_1}^{x_1} \partial_1u(t,x_2,\dots,x_n)\,dt\right|^pdx \\ &\le \int_\Omega \left(\int_{t=a_1}^{b_1} \left|\partial_1u(t,x_2,\dots,x_n)\right|\,dt\right)^pdx \\ &\le \int_\Omega \left(\int_{t=a_1}^{b_1} \left|\nabla u(t,x_2,\dots,x_n)\right|\,dt\right)^pdx \\ &\stackrel{(2)}{\le} \int_\Omega \left(\int_{t=a_1}^{b_1} 1\,dt\right)^{p/p'} \left(\int_{t=a_1}^{b_1} \left|\nabla u(t,x_2,\dots,x_n)\right|^p\,dt\right) dx \\ &\le (\text{diam}\,\Omega)^{p/p'} \int_\Omega\int_{t=a_1}^{b_1}|\nabla u(t,x_2,\ldots,x_n)|^p\,dtdx \\ &\stackrel{(3)}{\le} (\text{diam}\,\Omega)^{(p/p')+1} \int_\Omega|\nabla u(x_1,x_2,\ldots,x_n)|^p\,dx \\ &\stackrel{(4)}{=} (\text{diam}\,\Omega)^{p} \int_\Omega|\nabla u(x_1,x_2,\ldots,x_n)|^p\,dx \end{align} $$\therefore \|u\|_{L^p(U)} \le (\text{diam}\,\Omega)\,\|\nabla u\|_{L^p(U)}$$
$\stackrel{(1)}{=}$ is a result of $u\in W^{1,p}_0(U)$; $\stackrel{(2)}{\le}$ is a result Hölder's inequality; $\stackrel{(3)}{\le}$ is a result of Fubini's theorem; $\stackrel{(4)}{=}$ is a result of the definition of $p'$. Here, $C= \text{diam}\,\Omega$ so it only depends on $U$ but not on $p$ at all.
Is $C$ suppose to have some $p$-dependence that I happened to miss in the proof?