Indeed, the weak type estimate is useful. Using Fubini's theorem, we have
$$\int_E|f^{*}(x)|\mathrm dx=q\int_0^\infty t^{q-1}\lambda\{|f^*(x)|\chi_E\geqslant t\}\mathrm dt.$$
Notice that $$\lambda\{|f^*(x)|\chi_E\geqslant t\}\leqslant \min\left\{|E|;\frac{3^d}t\lVert f\rVert_{\mathbb L^1}\right\},$$
hence cut the integrals and conclude.
Theorem. Let $\mathcal{M}$ and $M$ denote the uncentered and centered Hardy-Littlewood maximal function using
balls, and let $\mathcal{M}_{c}$ and $M_{c}$ denote the uncentered and centered Hardy-Littlewood maximal
function using cubes. For $f\in L_{loc}^{1}(\mathbb{R}^{n})$,
$$\dfrac{2^{n}}{(v_{n}n^{n/2})}\leq\dfrac{M(f)}{M_{c}(f)}\leq\dfrac{2^{n}}{v_{n}}, \quad \dfrac{2^{n}}{(v_{n}n^{n/2})}\leq\dfrac{\mathcal{M}(f)}{\mathcal{M}_{c}(f)}\leq\dfrac{2^{n}}{v_{n}}\qquad\text{a.e.},$$
where $v_{n}$ denotes the volume of the unit ball in $\mathbb{R}^{n}$.
Proof. In what follows, $r>0$. Replacing $f$ by a translate, it suffices to establish the inequality at $x=0$. Observe that the cube $[-r,r]^{n}$ is almost everywhere contained in the open ball $B(0,n^{1/2}r)$. Whence,
$$\dfrac{1}{v_{n}n^{n/2}r^{n}}\int_{[-r,r]^{n}}\left|f\right|\leq\dfrac{1}{v_{n}n^{n/2}r^{n}}\int_{B(0,n^{1/2}r)}\left|f\right|=\dfrac{1}{\left|B(0,n^{1/2}r)\right|}\int_{B(0,n^{1/2}r)}\left|f\right|\leq Mf(0)$$
Multiplying by $1=2^{n}/2^{n}$ and taking the supremum over $r>0$ of the LHS, we obtain that
$$\dfrac{2^{n}}{v_{n}n^{n/2}}M_{c}f(0)\leq Mf(0)$$
Similarly, observe that the open ball $B(0,r)$ is contained in the cube $[-r,r]^{n}$. Whence,
$$\dfrac{1}{(2r)^{n}}\int_{B(0,r)}\left|f\right|\leq\dfrac{1}{(2r)^{n}}\int_{[-r,r]^{n}}\left|f\right|\leq M_{c}f(0)$$
Multiplying by $1=v_{n}/v_{n}$ and taking the supremum over $r>0$ of the LHS, we obtain that
$$\dfrac{v_{n}}{2^{n}}Mf(0)\leq M_{c}f(0)$$
A completely analogous argument establishes the inequality for the uncentered maximal functions. $\Box$
It's worth mentioning that these inequalities show that $\mathcal{M}_{c},\mathcal{M}$ are weak-type (1,1) operators and therefore are bounded operators $L^{p}(\mathbb{R}^{n})\rightarrow L^{p}(\mathbb{R}^{n})$ for $1<p\leq\infty$.
Best Answer
Here is a partial answer:
Let $Mf(x)=\sup_{r>0}\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f(y)|dy$ be the (centered) Hardy Littlewood Maximal Function.
Proof: Lower semicontinuity means $E(t)=\{x: Mf(x)>t\}$ is open for all $t\in \mathbb{R}$. Let $x\in E(t)$. Then $Mf(x)>t$, so there is an $r>0$ such that $Mf(x)\geq\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f(y)|dy>t$. If we choose $s>r$ small enough, then $$Mf(x)\geq\frac{1}{\lambda(B(x,r))}\int_{B(x,r)}|f(y)|dy>\frac{1}{\lambda(B(x,s))}\int_{B(x,r)}|f(y)|dy>t$$Then, for $z$ such that $|z-x|+r<s$, we will have $B(x,r)\subset B(z,s)$, so $$Mf(z)\geq\frac{1}{\lambda(B(z,s))}\int_{B(z,r)}|f(y)|dy>\frac{1}{\lambda(B(x,s))}\int_{B(x,r)}|f(y)|dy>t$$ Thus $z\in E(t)$, and the set of $z$ is open, so $E(t)$ is open.
Note: This is much easier to prove for the uncentered Hardy Littlewood Function.
Now, we are reduced to showing when $Mf$ is upper semicontinuous, since if a function is lower and upper semicontinuous somewhere, it is continuous. Searching hasn't gotten me much in terms of if and only if, but there is this result:
The proof is found in this article: Lemma 7.3. A slightly different requirement is the following:
The proof is in this paper (Lemma 3.4). Although $\chi_{[0,1]}$ is upper semicontinuous everywhere, $M\chi_{[0,1]}$ is not continuous at $0$ or $1$. The theorem holds because $$\chi_{[0,1]}(0)=1>M\chi_{[0,1]}(0)=\frac{1}{2}$$This provides a counterexample showing that we need the second condition.
Hope this helps as a partial answer!