Of course I do find dimensional analysis to be of great use, but I think your approach here is kind of cumbersome: dimensional analysis is useful if you can pick what the "interesting" dimensions in your problem are. Pulling out ampere(not totally sure what that ##A## means, but I guess it must be ampere), mass, time in the context of Ising model seems useless(at best) if not counterproductive. What I would care about here is just energy.
Since you have the solution, it should be clear to you that ##m## is a dimensionless quantity due to the fact that ##tanh##(and likewise all the transcendental functions) needs a dimensionless argument and returns a dimensionless value, as
@scottdave noted.
In case you didn't know the solution yet, you can tell that your field ##H## has the dimensions of an energy simply by looking at how the energy of the system is defined; you have
$$ E = -\sum_{i,k}J_{ik}\sigma_i\sigma_k + \sum_i\sigma_iH_i $$
where ##\sigma_i## is the "spin" of site ##i##, ##H_i## is the field at site ##i## and ##J_{ik}## is the coupling energy of sites ##i,k##.
Clearly you want the two addends to have the same dimensions; in principle you have an infinite number of possibilities of assigning the units(these are just conventions in the end) but given the units of spin, you can tell what the units of ##J## and ##H## are. Since usually the spins are introduced as dimensionless quantities, ## [\sigma]=1 ##, you can easily see that you must have ## [J] = [H] = [E] ## i.e. ##J## and ##H## both have dimensions of energy, making the first formula (dimensionally) correct. This, combined with the fact that the magnetization is defined as the expectation value of the spins, is consistent with ##[m]=1##.
In the end I guess that the point of my post is that no one likes to work with weird combinations of units, so just keep it as simple as possible.
