Integrating $\|y\|_2^{2-n}$ over $B_r(x)$ in $\mathbb{R}^n$ Generally for $n\geq3$ what is
$$\int_{B_r(x)}\frac{1}{\|y\|_2^{n-2}}dy$$
where $x,y\in\mathbb{R}^n$ and $\|x-y\|_2<r$?
Is the fact that $\|\cdot\|^{-1}$ (edit: $\|\cdot\|^{2-n}$) is harmonic on $\mathbb{R}^n\setminus \{0\}$ somehow useable?
 A: $\Vert\cdot\Vert^{2-n}$ is harmonic on $\mathbb R^n\setminus \{0\}$. (You wrote $\Vert\cdot\Vert^{-1}$, which is not usually harmonic, unless $n=3$ and then it's the same as $\Vert\cdot\Vert^{2-n}$.)
This means we may use the volume version of the mean value property:

If $u$ is harmonic on a closed ball then its value at the center of the ball is the volume integral average of its values inside the ball.

This follows from the regular mean value property (where the value at the center of the ball is the spherical area integral average of its values on the sphere), by performing a change of variables to polar coordinates.
Assuming $\Vert x \Vert_2 < r$, we have $y$ remaining separated from zero and there is no problem using the above. That is, dividing your integral by the volume of the ball $B_r(x)$ will give the value of the function $\Vert\cdot\Vert^{2-n}$ at the center of the ball, $x$.
If $\Vert x \Vert > r$ this is no longer true. To understand this, you have to go back to the proof of the mean value property. The volume version is just an "integrated" version of the usual version, which is $$\int_{\mathcal S^{n-1}} u(r \zeta) \,\mathrm d\sigma_{n-1}(\zeta) = u(0)$$ where $\sigma_{n-1}$ is the uniform measure on the unit sphere $\mathcal S^{n-1}$ normalized with total measure $1$, and we assume that $u$ is harmonic in the closed ball with center $0$ and radius $r$.
In our case, $u(y) = \Vert y - x \Vert^{2-n}$ and the above works as long as $r < \Vert x \Vert$ since in this range $u$ is harmonic. However, once we go to bigger values of $r$ and pass the singularity, we actually get
$$\int_{\mathcal S^{n-1}} u(r \zeta) \,\mathrm d\sigma_{n-1}(\zeta) = 0$$
instead of the value $u(0)$. (You should be able to figure out the volume integral from this.)
To see why this is zero, go back to the proof of the mean value property. It is usually proven by using Green's second identity with the function $u(y)$ (in our case, $\Vert y-x \Vert^{2-n}$ and the Newton kernel $v(y) = \Vert y \Vert^{2-n}$ in the domain $\Omega = \{ \varepsilon < \Vert y \Vert < r\}$ (taking out the singularity of the Newton kernel) and then letting $\varepsilon \to 0$.
In our case we need to take care of a second singularity, that of $u(y)$. So we use Green's second identity, this time in the domain $$\Omega = \{ \varepsilon < \Vert y \Vert < r ; \Vert y-x \Vert > \delta \}$$
and then let $\varepsilon \to 0$ and $\delta \to 0$. It turns out that the two values you picked up at the singularities cancel each other and you are left with zero.
