Such an estimate can be found in Grisvard "Elliptic problems in nonsmooth domains", Theorem 1.5.1.10.
It basically says
$$
\delta \|u\|_{L^2(\partial\Omega)}^2 \le \|\mu\|_{C^1(\bar\Omega)}
\left(\epsilon^{1/2}\|\nabla u\|_{L^2(\Omega)}^2
+ (1+\epsilon^{-1/2}) \|u\|_{L^2(\Omega)}^2
\right)
$$
for all $\epsilon\in(0,1)$, $u\in H^1(\Omega)$. The vector field $\mu$ has to be chosen to be $C^1(\bar\Omega,\mathbb R^n)$ such that
$$
\mu \cdot \nu \ge \delta
$$ on $\partial \Omega$ with $\nu$ the outer normal vector.
This could give you an estimate of the constant for a fixed domain at least. It should help to prove continuity with respect to domain variations as well.
It's easier to deal with a more general problem (in any case take $v=\nabla u$ and $f=\nu_i$ for each $i=1,\ldots,n$).
So your problem is basically this: given $v\in W^{1,p}(\partial \Omega)$ and $f:\partial \Omega \to \mathbb{R}$ a Lipschitz function, with $F:\bar{\Omega}\to \mathbb{R}$ an extension of $f$, is the following identity true?
$$
\gamma_0(v)f = \gamma_0(vF).\tag{1}
$$
Clearly this is true if $v\in C^\infty(\bar{\Omega})$, since in this case $\gamma_0(vF)= v|_{\partial \Omega} f$ and similarly for $u$ and $F$ (remember that $F\in W^{1,\infty}(\Omega)$). Now we have the following estimate
$$
\| vF \|_{W^{1,p}} \leq \| v\|_{W^{1,p}}\| F\|_{W^{1,\infty}}.
$$
Now take $v$ arbitrary and a sequence $v_k\in C^\infty(\bar{\Omega})$ with $v_k\to v$ in $W^{1,p}$. It follows from the estimate that $v_k F\to vF$ and, by continuity of the trace operator, that $\gamma_0(v_kF) \to \gamma_0(vF)$ in $W^{1-1/p,p}$. Again by continuity of the trace operator we have that $\gamma_0(v_k) f \to \gamma_0(v)f$ in $L^p(\partial \Omega)$. Collecting all this we obtain
$$
\gamma_0(v)f \overset{L^p(\partial \Omega)}{\leftarrow}\gamma_0(v_k)f=\gamma_0(v_kF) \overset{W^{1-1/p,p}}{\rightarrow} \gamma_0( vF)
$$
and (1) follows from this.
Also notice that if two functions $v_1,v_2 \in W^{1-1/p,p}$ satisfy $\| v_1-v_2\|_{L^p(\partial \Omega)}$ then $\| v_1-v_2\|_{W^{1-1/p,p}}=0$. This is because, by definition, $0\in W^{1,p}(\Omega)$ has $0$ trace so it's a representative of $v_1-v_2$.
Best Answer
The reason that the equality $Tu=u_{|\partial \Omega}$ is stated for functions in $C(\overline{\Omega})$ is that for other functions it is not clear what $u_{|\partial \Omega}$ means. Of course, we can take any function $u\in W^{1,p}(\Omega)$ and extend its domain to $\overline{\Omega}$ by letting $u$ be equal to $Tu$ on the boundary. This will be an instance of $Tu=u_{|\partial \Omega}$ being true despite $u$ not necessarily being in $C(\overline{\Omega})$. But this is merely a tautology that does not teach us anything new.
But there are useful and nontrivial results along the lines of your question: starting with $\phi\in L^p(\partial \Omega)$, one can extend $\phi$ to a function in $\Omega$ (by solving the Dirichlet problem for some elliptic operator) and then recover $\phi$ from $u$ via nontangential limits. This does not require $\phi$ to be continuous. For a simple example, take $\Omega=\{z\in\mathbb C: |z|<1\}$ and $u=\log|z-1|$. Then $u\in W^{1,p}(\Omega)$ for $p<2$. The trace operator agrees with the boundary values of $u$ understood in the sense of nontangential limits.