Question: I'm wondering if the minimal polynomials of $A, A^T\in\mathbb{M}\_{n\times n}(\mathbb{F})$ are somehow related? Are they the same?

If $m\_A(t)$ is the minimal polynomial of $A$, then $m\_A(A) = 0$. So, $(m\_A(A))^T = 0$. If two matrices $P,Q$ commute then $(PQ)^T = P^TQ^T$. Using this, we can see that $m\_A(A^T) =0$.

Is this enough to conclude that $m\_A(t)$ is also the **minimal polynomial** for $A^T$?

P.S.
Does this depend on whether or not the minimal polynomial splits over $\mathbb{F}$?

Answer: 
You don't need to consider any factorisation. Let $p(x)=\sum\_{i=1}^ka\_ix^i$ be any monic polynomial. If $p(A)=0$, then
$$
p(A^T)=\sum\_{i=1}^ka\_i(A^T)^i=\left(\sum\_{i=1}^ka\_iA^i\right)^T=p(A)^T=0.
$$
Since $A=(A^T)^T$, it is also true that $p(A^T)=0\Rightarrow p(A)=0$. Therefore $A$ and $A^T$ share the same set of annihilating polynomials. The monic polynomial of the lowest degree in this set is the common minimal polynomial of $A$ and $A^T$.
$m\_A$ and $m\_{A^T}$ are the minimal polynomials for $A$ and $A^T$ respectively.

So we have $m\_A(A) = 0$ and $m\_{A^T}(A^T) = 0$ by definition. Let $m\_A(t) = p\_1(t)p\_2(t)...p\_k(t)$ where $p\_i(t)\forall 1\leq i\leq k$ are the irreducible factors of $m\_A(t)$.

$m\_A(A) = 0 \implies (m\_A(A))^T = 0$, which is just $(p\_1(A)p\_2(A)...p\_k(A))^T = 0$. Since polynomials in $A$ commute, this is just $(p\_1(A))^T(p\_2(A))^T...(p\_k(A))^T = 0 \implies p\_1(A^T)p\_2(A^T)...p\_k(A^T) = 0$. So $m\_A(A^T) = 0$, which means that $m\_{A^T}\vert m\_A$.

Starting with $m\_{A^T}(A^T) = 0$, we can similarly prove that $m\_A\vert m\_{A^T}$.

So $m\_A(t) = m\_{A^T}(t)$