I'm looking for an upper bound on $\|\partial_{ij} u\|_{L^\infty}$, where $u$ is the solution to $$ \Delta u = -1 \quad \text{in} \quad \Omega,\\ u = 0 \quad \text{on} \quad \partial\Omega. $$ The upper bound may depend on the diameter of $\Omega$, but not on other properties of $\Omega$.
I think the question may have something to do with "Schauder estimates" but I'm struggling to make sense of them. Would be grateful for pointers to easy-to-read resources on this topic.
Let $R= \mathrm{diam} \,(\Omega)$ and, after possibly translating $\Omega$, we can assume that that $\Omega \subset B_R$. Let $v = u -\frac1 {2n}(R^2- \vert x \vert^2 )$. Then $$\Delta v=0 \qquad \text{in } \Omega.$$ By the Cauchy estimates, for each $\Omega'\subset \subset \Omega$, $$\| D^2 v \|_{L^\infty(\Omega')} \leqslant C \| v \|_{L^\infty(\Omega)} \tag{$\ast$}$$ where $C=C(n,\mathrm{diam} (\Omega'),\mathrm{dist}\,(\partial \Omega,\Omega'))>0.$ Since $D_{ij}v=D_{ij}u+\frac 1 n \delta_{ij}$, it follows from $(\ast)$ that $$\| D^2 u \|_{L^\infty(\Omega')} \leqslant C \big ( \| u \|_{L^\infty(\Omega)} +1 \big ). \tag{$\ast\ast$}$$
Now let us estimate $\| u \|_{L^\infty(\Omega)}$. Since $u$ is superharmonic and zero on $\partial \Omega$ the comparison principles says that $u\geqslant 0$. Moreover, $v$ is harmonic and $v\leqslant 0$ on $\partial \Omega$, so the comparison principle tells us that $v\leqslant 0$ in $\Omega$, that is, $$u\leqslant \frac1 {2n}(R^2- \vert x \vert^2 )\leqslant \frac 1 {2n} (\mathrm {diam} \,\Omega)^2. $$ Thus, combining this with $(\ast\ast)$, we have that for each $\Omega'\subset \subset \Omega$ $$\| D^2 u \|_{L^\infty(\Omega')} \leqslant C (\mathrm {diam}\, \Omega)^2. $$
Note: If you would like estimates up to the boundary then we need information about the regularity of the boundary of $\Omega$.
Edit: To answer you question in the comments about the constant in the Cauchy estimate. I don't know of a specific textbook which has exactly what you are looking for (often they just say $C$ depends on $\Omega$ and $\Omega'$). Here's what I was thinking: The standard Cauchy estimates say that $$ \| D^2 u \|_{L^\infty(B_r(x))} \leqslant \frac{C}{r^{n+2}}\|u\|_{L^1(B_r(x))}\leqslant\frac{C}{r^2}\|u\|_{L^\infty(B_r(x))}$$ for each $B_r(x) \subset \Omega$ and $C=C(n)$. By compactness of $\Omega'$ in $\Omega$, let $N$ be the smallest integer such that there exists $\{x_j\}_{j=1}^N\subset \Omega'$ such that $$ \Omega'\subset \bigcup_{j=1}^N B_r(x_j)$$ where $r=\mathrm{dist}\,(\partial \Omega,\Omega')/2$. Note that we have that the right hand side is a subset of $\Omega$ by our choice of $r$. Then $$\| D^2 u \|_{L^\infty(\Omega')} \leqslant \sum_{j=1}^N\| D^2 u \|_{L^\infty(B_r(x))}\leqslant \frac C {r^2}\sum_{j=1}^N\|u\|_{L^\infty(B_r(x))}\leqslant \frac C {r^2}N\|u\|_{L^\infty(\Omega)}.$$ To estimate $N$ note that $\Omega' \subset B_{\mathrm{diam}(\Omega')/2}(x)$ for any $x\in \Omega'$ and that $B_{\mathrm{diam}(\Omega')/2}(x)$ can be covered with $\approx \mathrm{diam}(\Omega')^n / \mathrm{dist}\,(\partial \Omega,\Omega')^n$ (you would need to make this rigorous) balls of radius $\mathrm{dist}\,(\partial \Omega,\Omega')/2$. Hence, you should get $N \leqslant C\mathrm{diam}(\Omega')^n / \mathrm{dist}\,(\partial \Omega,\Omega')^n$ for some universal constant $C$.