How to determine $\lim_{\epsilon \to 0} (\sup_{z \in \partial B(x,\epsilon)} |G(x,z)|$) for $n \geq 3$?
I know that $G(x,z) = \phi (z-x) - w^{x}(z)$.
And for $n \geq 3$ the fundamental solution is given by: $\hspace{2mm}$ $\phi(z-x) = \frac{1}{(n-2)w_{n}} \cdot \frac{1}{|z-x|^{n-2}} = \frac{1}{(n-2)w_{n}} \cdot \frac{1}{\epsilon^{n-2}}$
Now I also want to express the corrector function $w^{x}(z)$ in terms of $\epsilon$, but I don't know how. Or is the corrector function bounded on $\partial B(x,\epsilon)$?
In the lecture notes I can only find that: $w^{x}(z) = \phi(\frac{|x|}{r}z - \frac{r}{|x|}x)$ for $x \neq 0$, $x \neq z$ and $w^{x}(z) = \frac{1}{(n-2)w_{n}} \cdot \frac{1}{r^{n-2}}$ for $x = 0$ ($n \geq 3$). But here $z$ is in $\partial B_{r}(0)$. And I want that $z \in \partial B(x, \epsilon)$.
The function $w^x(\cdot)$ is bounded on the domain you are working on $\bar{\Omega}$ (the bound will depend on $x$). This follows from the maximum principle: $w^x$ solves the PDE $$ \begin{cases} \Delta w^x = 0 & \text{ on } \Omega\\ w^x(z) = \phi(z - x) & z \in \partial \Omega, \end{cases} $$ and $\phi(\cdot - x)$ is an explicit smooth function.
In the case of $\Omega$ a ball (which appears to be where you formulated the question) you actually can obtain an explicit formula for $w^x$: $w^x(z) = - C|x|^{2-n}\phi(z - x^*)$ where $x^*$ is the inversion of the point $x$, and in particular lies outside $\Omega$; this function is smooth on $\bar{\Omega}$.