Equivalence of two formulation of Maxwell equations on manifolds
Maxwell's equations in differential form notation reads $$ dF = 0~, \qquad d\ast F = \ast J~. $$ We now show that these equations are equivalent to $\nabla_{[a} F_{bc]} = 0$, $\nabla_a F^{ab} = J^b$.
First, by definition \begin{align} \nabla_{[a} F_{bc]} = \partial_{[a} F_{bc]} + \Gamma^d_{[ab}F_{c]d} - \Gamma^d_{[ac} F_{b]d} \end{align} If the connection is torsion-free (not necessarily the Levi-Civita connection) then $\Gamma^a_{[bc]} = 0$ so that \begin{align} \nabla_{[a} F_{bc]} = \partial_{[a} F_{bc]} = \frac{1}{3} (dF)_{abc} \end{align} The last equality is true by definition. Thus, $\nabla_{[a} F_{bc]} \implies dF = 0$.
Next, consider the second equation \begin{align} \nabla_a F^{ab} = \partial_a F^{ab} + \Gamma^a_{ac} F^{cb} + \Gamma^b_{ac} F^{ac} \end{align} Again, if the connection is torsion free, then the last term is zero. To simplify the second term, we have to assume that $\Gamma$ is the Levi-Civita connection so that $$ \Gamma^a_{ac} = \frac{1}{2} g^{ab} ( \partial_a g_{cb} + \partial_c g_{ab} - \partial_b g_{ac} ) = \frac{1}{2} g^{ab} \partial_c g_{ab} = \frac{1}{2} \partial_c \log \det g = \frac{1}{\sqrt{\det g}}\partial_c \sqrt{\det g} $$ Then, we have \begin{align} \nabla_a F^{ab} = \partial_a F^{ab} + \frac{1}{\sqrt{\det g}}\partial_c \sqrt{\det g} F^{cb} = \frac{1}{\sqrt{\det g}} \partial_a \left( \sqrt{\det g} F^{ab} \right)~. \end{align} So we can write Maxwell's equation as $$ \partial_e \left( \sqrt{\det g} F^{ed} \right) = \sqrt{\det g} J^d $$ Now, contract both sides with the Levi-Civita symbol (not tensor), ${\tilde \varepsilon}_{abcd}$ to get $$ \partial_e \left( \sqrt{\det g} {\tilde \varepsilon}_{abcd} F^{ed} \right) = \sqrt{\det g}{\tilde \varepsilon}_{abcd} J^d $$ Now, recall the the Levi-Civita tensor is ${\varepsilon}_{abcd} = \sqrt{\det g} {\tilde \varepsilon}_{abcd}$ $$ \partial_e \left( {\varepsilon}_{abcd} F^{ed} \right) = {\varepsilon}_{abcd} J^d = (\ast J)_{abc} \tag{1}$$ The last equality is true by definition. Finally, we wish to write the LHS in terms of $\ast F$. To do this, we write $$ F^{ed} = -\frac{1}{2} \varepsilon^{edmn} (\ast F)_{mn} $$ Then, $$ (\ast J)_{abc} = \frac{1}{2} \partial_e \left( {\varepsilon}_{abcd}\varepsilon^{demn} (\ast F)_{mn} \right) $$ Then, we use the property $$ {\varepsilon}_{abcd}\varepsilon^{demn} = 6 \delta^e_{[a} \delta^m_b \delta^n_{c]} $$ Finally, $$ (\ast J)_{abc} = 3 \delta^e_{[a} \delta^m_b \delta^n_{c]} \partial_e (\ast F)_{mn} = 3 \partial_{[a} (\ast F)_{bc]} = ( d \ast F )_{abc} $$ where again, the last equility is the definition of $d$. Thus, we see that $\nabla_a F^{ab} = J^b \implies d \ast F = \ast J$.
QED.
PS - To answer your last question. The differential form notation is aware of the connection through the Hodge dual in which $\sqrt{\det g}$ enters. Note also that in the divergence of any $p$-form, only the following component of the connection appears - $\Gamma^a_{ab}$ which depends entirely on $\sqrt{\det g}$. Other components never appear, i.e. in full generality $$ \nabla_a T^{[abc\cdots]} = \frac{1}{\sqrt{\det g}} \partial_a \left( \sqrt{\det g} T^{[abc\cdots]} \right)~. $$
The two formulations (1) & (2) are equivalent, mainly because:
the covariant and the partial derivative of an antisymmetric $(0,2)$ tensor $F_{\mu\nu}$ is equivalent for a torsion-free connection.
the Levi-Civita connection preserves the metric $\nabla_{\lambda} g_{\mu\nu}=0$.
OP's eq. (2a) reads in local coordinates $$ \pm J^{\nu}~=~\nabla_{\mu} F^{\mu\nu}~\equiv~\partial_{\mu} +\Gamma_{\mu\lambda}^{\mu}F^{\lambda\nu} +\Gamma_{\mu\lambda}^{\nu}F^{\mu\lambda}~=~\frac{1}{\sqrt{|g|}} \partial_{\mu}(\sqrt{|g|} F^{\mu\nu})~=~(\delta F)^{\nu} \tag{2a}$$ in Minkowski signature $(\pm, \mp,\mp,\mp)$. Here $\delta$ is the Hodge co-differential, up to sign conventions.
If $\nabla$ is a torsion-free connection (not necessarily Levi-Civita), then for an arbitrary $k$-form $\omega_{a_1...a_k}$ we have $(d\omega)_{a_1...a_{k+1}}=(k+1)\partial_{[a_1}\omega_{a_2...a_{k+1}]}=(k+1)\nabla_{[a_1}\omega_{a_2...a_{k+1}]}$. If you expand in terms of connection coefficients, the symmetric connection coefficients will be killed by the antisymmetrization.
So $\nabla_{[a} F_{bc]}=0\Leftrightarrow dF=0$.
For the first relation $\nabla_a F^{ab}=J^b$, look up the codifferential. It is defined as $\delta\omega=(-1)^k\star^{-1}d\star\omega$, so it is basically $\delta=\pm\star d\star$ with the usual sign clusterf*ck one has to deal with when using the Hodge star. The codifferential is in some sense the "adjoint" operator of $d$, and it decreases the degree of a differential form by one, and also knows $\delta\delta=0$ (and I think its own version of the Poincaré-lemma as well).
It can be shown (see for example General Relativity by Norbert Straumann), that the codifferential acts on a differential form by taking its divergence via the Levi-Civita connection (remember that the Hodge star and so the codifferential requires a metric, so it also "sees" the Levi-Civita connection), so $(\delta\omega)^{a_1...a_{k-1}}=\pm\nabla_a\omega^{aa_1...a_{k-1}}$ (once again annoying signs).
Now, the Maxwell equation in question is given by $d\star F=\mathcal{J}$, but then $\mathcal{J}$ is a 3-form here. So let us instead have $\mathcal{J}=\star J$, then $$ d\star F=\star J, \\ \star^{-1} d\star F=J=\delta F. $$ But by the previous discussion $(\delta F)^b=\pm\nabla_aF^{ab}$, so your formula is given.