What is the equivalent to $Box A^alpha =- mu_0 J^alpha$ using index-free notation from differential geometry?

Maxwell's equations can expressed in the language of differential forms.

In terms of tensors, Maxwell's equations can be written as

$$ nabla_{nu}F^{munu} = mu_0J^{mu},, nabla_{[alpha}F_{munu]} = 0 $$ where $F_{munu}$ is the Faraday tensor.

In terms of differential forms, these equations are written as

$$ d{bf F} = 0, d{star}{bf F} = mu_0{bf J},, $$ where ${bf F} = frac{1}{2}F_{munu}dx^{mu} wedge dx^{nu},$, $star$ is the Hodge star operator and ${bf J} = frac{1}{3!}{cal J}_{alphabetasigma}dx^{alpha}wedge dx^{beta}wedge dx^{sigma}$ is the 3-form associated with the current density four-vector. The components of ${cal J}_{alphabetasigma}$ are related to those of the current density 4-vector.

The Faraday 2-form, ${bf F}$, satisfies, ${bf F} = d{bf A}$, where ${bf A} = A_{mu}dx^{mu},.$

See https://en.wikipedia.org/wiki/Maxwell%27s_equations

and

https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field

Note that there are different sign conventions so be careful when consulting different websites/books on this subject.

If $$delta equiv *d*$$ then the Maxwell equations amount to $$delta d A = J$$ with $mu_0 =1$.

Starring the above equation, you get conservation of current. Taking the dual twice will get you to the form you started with, that is modulo some signs. Taking the dual of the above equation you get conservation of current $$ *delta d A sim d*dA = *J $$ now apply $d$ and get as a consistency on the equations of motion current conservation: $$ d*J = 0. $$

It is interesting exercise to consider the action from which those equations of motion come from, and discover how the variational derivative would be expressed in terms of differential forms. My suggestion, without the current, would be

$$ S_{EM} = int dA wedge * dA $$