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 $1<p\leq\infty$.
Here is a little refinement of your sketch of proof.
A function $f$ on $\mathbb{R}^n$ is called locally integrable, denoted $f\in L^1_{loc}(\mathbb{R}^d,\lambda_d)$, if $f\mathbb{1}_E\in\mathcal{L}_1(\mathbb{R}^d,\lambda_d)$ for all bounded measurable set $E$.
Claim: For such function $f$ and fixed $r>0$, the map $$M_rf:x\mapsto \frac{1}{\lambda(B(x;r))}\int_{B(x;r)}f\,d\lambda=\frac{1}{\omega_n r^n}\int_{B(x;r)}f\,d\lambda$$
is continuous and hence measurable.
Indeed, for fixed $x\in\mathbb{R}^d$ we have $f\mathbb{1}_{B(x;2r)}\in L^1$. Since $\|\mathbb{1}_{B(y;r)}-\mathbb{1}_{B(x;r)}\|_1\xrightarrow{y\rightarrow x}0$, the conclsion follows from
\begin{align*}
\Big|\int_{B(y;r)}f\,d\lambda-\int_{B(x;r)}f\,d\lambda\Big|\leq \int_{B(x;r)\triangle B(y;r)}|f\mathbb{1}_{B(x;2r)}|\,d\lambda.
\end{align*}
for $|x-y|<r$. Continuity follows from the following basic result:
$f\in L_1(\mu)$ then for any $\varepsilon>0$, there is $\delta>0$ such that $\mu(A)<\delta$ implies that $\int_A|f|\,d\mu<\varepsilon$.
It follows that the map
$x\mapsto\sup_{r>0}\frac{1}{\lambda(B(x;r))}\int_{B(x;r)}\,d\lambda=\sup_{r>0}M_rf(x)$ is lower semicontinuous and so, measurable.
Best Answer
To determine in which $L^p$ spaces a function $g$ belongs, the behavior of $\lambda(\alpha)$ for large $\alpha is pertinent.
If $0=\alpha_0<\alpha_1<\alpha_2<\ldots$ is a partition of $[0,\infty)$, then $$\int_{\mathbb{R}^{n}}|g(y)|^{p}dy = \sum_{k=1}^\infty \int_{y: \alpha_ {k-1} <|g(y)| \le \alpha_k} |g(y)|^{p}dy \,,$$ so $$\sum_{k=1}^\infty \alpha_{k-1}^p \Bigl(\lambda(\alpha_ {k-1})- \lambda(\alpha_k) \Bigr) \le \int_{\mathbb{R}^{n}}|g(y)|^{p}dy \le \sum_{k=1}^\infty \alpha_{k}^p \Bigl(\lambda(\alpha_ {k-1})- \lambda(\alpha_k)\Bigr) \,,$$
Passing to the limit as the partition is refined gives the desired relation.