Here is an argument that MLC is equivalent to an arithmetic statement. Let $M$ denote the Mandelbrot set, and let $\mathbb{C}_{\mathbb{Q}}$ denote the set of rational complex numbers. First note that by a fairly standard argument, the set of pairs $\{(z,\varepsilon)\in \mathbb{C}_{\mathbb{Q}}\times \mathbb{Q}: B_{\leq \varepsilon}(z)\cap M = \varnothing\}$, where $B_{\leq \varepsilon}(z)$ is the closed ball of radius $\varepsilon$ centered at $z$, is computably enumerable. This means that there is a uniform procedure for computing an element of $B_{< 2\varepsilon}(z)\cap M$ (for $z \in \mathbb{C}_{\mathbb{Q}}$ and $\varepsilon \in \mathbb{Q}$ with $B_{\leq \varepsilon} \cap M \neq \varnothing$) from $0'$, and therefore $0'$ uniformly computes a countable dense sequence $(m_i)_{i<\omega}$ of elements of $M$. (It is highly likely that there is a uniformly computable dense sequence in $M$, but I do not see an argument.) Let $M_0$ denote the set $\{m_i:i<\omega\}$.
Say that a set $F$ in a metric space $X$ is $\delta$-connected if for any $x,y \in F$, there is a finite chain $z_0,z_1,\dots,z_{n}$ with $z_0 = x$, $z_n = y$, and $d(z_i,z_{i+1}) < \delta$ for each $i<n$. Note that in a compact metric space, a closed set is connected if and only if it is $\delta$-connected for every $\delta > 0$. (Furthermore, recall that $M$ is a compact metric space.)
I claim that MLC is equivalent to the following arithmetic statement:
For every rational $\varepsilon > 0$, there is an $n < \omega$ such that for every rational $\delta >0$, there are finite sets $F_0,F_1,\dots,F_{n-1} \subset M_0$ such that
- each $F_i$ has diameter less than $\varepsilon$,
- each $F_i$ is $\delta$-connected, and
- $M$ is covered by $\bigcup_{i<n} \bigcup_{x \in F_i} B_{<\delta}(x)$.
First note that this is indeed arithmetic. Each of the bulleted conditions is $0'$-semi-decidable and therefore $0''$-decidable. (The last one relies on the fact that the complement of $M$ in $\mathbb{C}$ is computably enumerable.) This means that ostensibly the bullet points each have complexity $\Sigma^0_2$. Since the quantifiers out front are $\forall \exists \forall \exists$, this implies that the complexity of the entire statement is $\Pi^0_5$. If we knew there was a uniformly computable dense sequence in $M$, I think this shows that it is $\Pi^0_4$. Also note that there's nothing particularly special about the Mandelbrot set in all of this. All we're really using is that it's a compact co-c.e. subset of a computable metric space (where I mean co-c.e. in the sense of computable analysis).
Let's see that MLC implies the statement. Assume that $M$ is locally connected. Fix a rational $\varepsilon > 0$. Find a finite open cover $U_0,\dots,U_{n-1}$ of $M$ such that each $U_i$ is connected and has diameter less than $\frac{1}{2}\varepsilon$. Let $G_i$ be the closure of $U_i$ for each $i<n$. Note that the diameter for $G_i$ is at most $\frac{1}{2}\varepsilon < \varepsilon$. Now fix a rational $\delta > 0$. Since each $G_i$ is also connected, we can find a finite subset $H_i$ of $G_i$ such that $G_i \subseteq \bigcup_{x \in H_i} B_{<\frac{1}{2}\delta}(x)$. Since $G_i$ is connected, it is $\frac{1}{2}\delta$-connected and we can find a finite set $E_i \subseteq G_i$ with $H_i \subseteq E_i$ such that $E_i$ is $\frac{1}{2}\delta$-connected. (Just add a chain between each pair of elements of $H_i$.) $E_i$ still covers $G_i$ in the same way, so $M \subseteq \bigcup_{i<n} \bigcup_{x \in E_i} B_{<\frac{1}{2}\delta}(x)$. Finally, since $E_i \subseteq G_i$, the diameter of $E_i$ is less than $\frac{1}{2}\varepsilon$. Now finally, for each $x \in E_i$, find $m_x \in M_0$ such that $d(x,m_x) < \min\left(\frac{1}{3}\varepsilon, \frac{1}{5}\delta\right)$, and let $F_i = \{m_x : x \in E_i\}$. It is immediate that the diameter of $F_i$ is less than $\frac{1}{2}\varepsilon + \frac{1}{3}\varepsilon < \varepsilon$. Furthermore, $G_i \subseteq \bigcup_{x \in F_i} B_{< \delta}(x)$ and each $F_i$ is $\delta$-connected by the triangle inequality (since $\frac{1}{5}\delta + \frac{1}{2}\delta + \frac{1}{5}\delta < \delta$).
Now let's see that the arithmetic statement implies MLC. We'll need a small lemma.
Lemma. A compact metric space $X$ is locally connected if and only if for every rational $\varepsilon > 0$, there is a finite closed covering $G_0,G_1,\dots,G_{n-1}$ of $X$ such that each $G_i$ is connected and has diameter at most $\varepsilon > 0$.
Proof. The forward direction is obvious. For the reverse direction, suppose that the second statement is true. Fix $\varepsilon > 0$ and find a finite covering $G_0,G_1,\dots,G_{n-1}$ of $X$ such that each $G_i$ is connected and has diameter at most $\frac{1}{2}\varepsilon > 0$. For any $x \in X$, let $G_x = \bigcup\{G_i:i<n,~x \in G_i\}$. Note that since the $G_i$'s form a closed cover, we have that $x$ is in the interior of $G_x$. Furthermore, $G_x$ is connected (since it is a union of connected sets that contain a common point) and has diameter at most $\varepsilon$. Since we can do this for any $x \in X$ and any rational $\varepsilon > 0$, we have that $X$ is connected im kleinen at every point, which implies local connectedness (in any topological space). $\square_{\text{Lemma}}$
Fix a rational $\varepsilon > 0$. By assumption, there is an $n$ such that for every rational $\delta >0$, there are finite sets $F_0,F_1,\dots,F_{n-1} \subset M_0$ satisfying the stated properties above. For each $k < \omega$, let $F^k_0,F^k_1,\dots,F^k_{n-1}$ be a particular choice of such sets for $\delta = 2^{-k}$. Since the space of closed subsets of $M$ is compact under the Hausdorff metric, we can find an increasing sequence $(k(j))_{j<\omega}$ of natural numbers such that $(F^{k(j)}_i)_{j<\omega}$ converges in the Hausdorff metric for each $i<n$. Let $G_i = \lim_{j \to \infty} F^{k(j)}_i$. Fairly easy arguments show that each $G_i$ has diameter at most $\varepsilon$, each $G_i$ is $\delta$-connected for every $\delta > 0$ and therefore is connected, and $M$ is covered by $\bigcup_{i <n} \bigcup_{x \in G_i} B_{<\delta}(G_i)$ for every $\delta > 0$ and so $M = \bigcup_{i<n} G_i$. Therefore, by the lemma, $M$ is locally connected.
