If $u$ is a harmonic function, then $|D^\alpha u(x)| \le c\|u\|_{L^1(B(x, r))}/r^{n + |\alpha|}$ I want to show the following: Let $u \in C^2(\Omega)$ be a harmonic function and $\overline{B(x, r)} \subset \Omega \subset \mathbb R^n$ for some $x \in \Omega$ and $r > 0$. For $\alpha \in \mathbb N^n$ a multi-index, I want to show that
$$|D^\alpha u| \le \frac{c}{r^{n + |\alpha|}} \|u\|_{L^1(B(x, r))},$$
for $c = c(\alpha, n) > 0.$ I have absolutely no idea how to do that.. Obviously, by induction, it is enough to show the statement for $\alpha = (1, 0, \ldots, 0)$, and I applied Stokes' theorem to get rid of the derivative, which gave me
\begin{align}
|\partial_1 u(x)| &= \left|\frac{c}{r^n}\int_{B(x, r)}\partial_1 u(y)dy\right|\\
&= \left|\frac{c}{r^n}\int_{\partial B(x, r)} u(z) \nu_1d\sigma(z)\right|\\
&\le \frac{c}{r^n}\|u\|_{L^1(\partial B(x, r))},
\end{align}
but I wasn't able to get any further than that. Any help is welcome.
 A: Your work is totally fine, but I don't think it can obtain what you desire to derive. (Observe that the power of $R$ in the denominator is $n$, instead of $n+1$, for example.)
To prove it, we start with the 1-st derivative. We use the mean value formula on $B\left(x,\frac{r}{2}\right)$ instead.
Note that
$$ \left|\partial_i u\left(x\right)\right|=\frac{2n}{r}\left\|u\right\|_{L^{\infty}\left(\partial B\left(x,\frac{r}{2}\right)\right)}=\frac{2n}{r}\left|u\left(p\right)\right|$$
for some $p\in \partial B\left(x,\frac{r}{2}\right)$.
Now, note that $B\left(p,\frac{r}{2}\right)\subset B\left(x,r\right)\subset \Omega$ (This is crucial), so we have
$$ \left|u\left(p\right)\right|\leq \frac{2^n}{\omega_nr^n}\left\|u\right\|_{L^1\left(B\left(p,\frac{r}{2}\right)\right)}\leq  \frac{2^n}{\omega_nr^n}\left\|u\right\|_{L^1\left(B\left(x,r\right)\right)},$$
where $\omega_n$ is the volume of the unit $n$-ball. Thus,
$$ \left|\partial_i u\left(x\right)\right| \leq \frac{2n}{r}\cdot \frac{2^n}{\omega_nr^n}\left\|u\right\|_{L^1\left(B\left(x,r\right)\right)}=\frac{c}{r^{n+1}}\left\|u\right\|_{L^1\left(B\left(x,r\right)\right)}.$$
For the rest of derivatives, you can use inductive argument to derive, and the important step is you need to use the smaller ball so that you can  obtain the result by combining the estimate data on the boundary, as it is shown above. ($B\left(x,\frac{r}{k}\right)$ would do the job if you want to derive the $k$-th derivative estimate from $k-1$-th, but you should be aware of the choice of balls of the boundary point; namely, you cannot take a ball so large that it exceeds $\Omega$.)
