Let $f^*(\mathbf x) = \sup \frac{1}{Q} \int_Q |f(\mathbf y)|d\mathbf y$. Sorry, my questions actually were:
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Since $\chi_E^*(x) \ge \frac{C}{\lvert x \rvert^n}$ for large $x$, then using the tranformation formula (or in this case integration in polar coordinates), we have $$\int_{\{\lvert x \rvert \ge R\}} \chi_E^*(x) dx \ge \int_{\{\lvert x \rvert \ge R\}} \frac{C}{\lvert x \rvert^n}dx = \alpha_{n} C \int_R^\infty \frac{r^{n-1}dr}{r^n} = \alpha_n C \int^\infty_{R} \frac{dr}{r} = \infty,$$ where $\alpha_n$ is the surface measure of $S^{n-1} = \{x \in \mathbb R^n \, : \| x \|_2 = 1\}$. This shows that $\chi_E^*$ is not integrable.
To prove the inequality, if $E$ is bounded, then for $x$ large, $x \not \in E$, and the distance $\text{dist}(x,E) \sim \lvert x \rvert$ as $\lvert x \rvert \to \infty$ (do you see why this is? A picture may help). Now if the cube $Q^x$ contains $E$, it has side length greater than some constant times $\text{dist}(x,E)$ (indeed, the cube centered at $x$ of side length $2\text{dist}(x,E)/\sqrt n$ touches $E$ but does not contain any of the interior of $E$). Thus we will have $\lvert Q^x \rvert \sim \text{dist}(x,E)^n \sim \lvert x \rvert^n$.
If $E$ is bounded, let $M=\sup_{y\in E}|y|$. If $|x|$ is much larger than $M$, then for a cube centered at $x$ to intersect $E$ at all, the cube must have side length at least $c|x|$ for some constant $c$ (this constant is somewhat annoying to write down explicitly; if we were dealing with a ball instead of a cube we could just say the radius must be at least $|x|-M$ which is greater than $|x|/2$ for $|x|$ large, but for a cube we need an additional constant relating the $\ell^\infty$ norm to the Euclidean norm). So for large $|x|$, we have $\chi_E^*(x)\leq \frac{|E|}{c^n|x|^n}$ since if $Q$ is any cube centered at $x$ which intersects $E$ we have $|Q\cap E|\leq|E|$ and $|Q|\geq c^n|x|^n$. So we can take $c_2=1/c^n$.
On the other hand, though, there is another constant $c'$ such that a cube of side length $c'|x|$ centered at $x$ will contain all of $E$. (Again, if we were taking a ball instead of a cube we could take a ball of radius $|x|+M$ which is at most $2|x|$ for $|x|$ large.) This cube has measure $c'^n|x|^n$ and witnesses that $\chi_E^*(x)\geq \frac{|E|}{c'^n|x|^n}$, so we can take $c_1=1/c'^n$.
In particular, if $|E|>0$, then this gives a constant $b>0$ such that $\chi_E^*(x)>\frac{b}{|x|^n}$ when $|x|$ is sufficiently large. Since the integral of $\frac{1}{|x|^n}$ over $\mathbb{R}^n$ diverges (even away from the singularity at the origin), this means that that the integral of $\chi_E^*$ diverges too.