(Friedrichs' Inequality): $$ \|u-\bar{u}\|_{W_{p}^{1}(\Omega)} \leq C|u|_{W_{p}^{1}(\Omega)} $$ where $\bar{u}=\frac{1}{|\Omega|} \int_{\Omega} u(x) d x$.
I'v learnt some proofs about this inequality like the application of normed-equivalence theorem, but yesterday I find another proof which I think is strange(using Bramble-Hilbert).
I even think the proof in that book is wrong
You just need to pay attention to the conclusion of the next lemma and $Q^{m} u$ is actually a polynomial of degree $m - 1$. The next two lemma is copied from the book.
Lemma: (Bramble-Hilbert) Let $B$ be a ball in $\Omega$ such that $\Omega$ is star-shaped with respect to $B$ and such that its radius $\rho>(1 / 2) \rho_{\max } .$ Let $Q^{m} u$ be the Taylor polynomial of order $m$ of $u$ averaged over $B$ where $u \in W_{p}^{m}(\Omega)$ and $p \geq 1$. Then $$ \left|u-Q^{m} u\right|_{W_{p}^{k}(\Omega)} \leq C_{m, n, \gamma} d^{m-k}|u|_{W_{p}^{m}(\Omega)} \quad k=0,1, \ldots, m, $$ where $d=\operatorname{diam}(\Omega)$.
(Friedrichs' Inequality) Suppose $\Omega$ is star-shaped with respect to a ball $B$. Then for all $u \in W_{p}^{1}(\Omega)$, $$ \|u-\bar{u}\|_{W_{p}^{1}(\Omega)} \leq C_{n, \gamma}|u|_{W_{p}^{1}(\Omega)} $$ where $\bar{u}=\frac{1}{|\Omega|} \int_{\Omega} u(x) d x$.
Proof. Observe that, by Hölder's inequality, $$ \|\bar{\phi}\|_{L^{p}(\Omega)} \leq\|\phi\|_{L^{p}(\Omega)} \quad \forall \phi \in L^{p}(\Omega) . $$ Therefore we have, for any constant $c$, $$ \|u-\bar{u}\|_{L^{p}(\Omega)}=\|(u-c)-\overline{(u-c)}\|_{L^{p}(\Omega)} \leq 2\|u-c\|_{L^{p}(\Omega)} $$ and hence, by using the lemma above, $$ \|u-\bar{u}\|_{L^{p}(\Omega)} \leq 2 \inf _{c \in \mathbb{R}}\|u-c\|_{L^{p}(\Omega)} \leq C_{*}|u|_{W_{p}^{1}(\Omega)} $$ where the constant $C_{*}$ depends only on the dimension $n$ and the chunkiness constant $\gamma$ of the domain. Consequently the inequality holds with $C_{n, \gamma}=1+C_{*} .$
Question: Why $C_{*}$ is independent on d?
You are correct in that $C_*$ is not independent of $d$ in general, which can be seen by a scaling argument. Consider any non-trivial $u \in W^{1,p}_0(B(0,1))$ and let $u_r(x) = u(rx) \in W^{1,p}_0(B(0,r))$ where $B(x,r)$ denotes the open ball centred at $x$ with radius $r.$ Then $$ \lVert u_r\rVert_{L^p(B(0,r))} = r^{-\frac np} \lVert u \rVert_{L^p(B(0,1))},\ \text{ while } \ \lVert \nabla u_r\rVert_{L^p(B(0,r))} = r^{1-\frac np} \lVert \nabla u \rVert_{L^p(B(0,1))}.$$ Since $r>0$ is arbitrary, no such estimate can hold independently of $r.$
Note you didn't define $\gamma,$ but I assume the associated constant for $B(0,r)$ is independent of $r.$
Also it's worth pointing out that if $p < n,$ we have the Poincaré-Sobolev inequality which states that $$ \lVert u -\overline u \rVert_{L^{p^*}(\Omega)} \leq C_{n,\gamma} \lVert \nabla u \rVert_{L^p(\Omega)}, $$ where $p^* = \frac{np}{n-p}.$