You are asking about a double-well potential where the two wells are separated by a finite square wall. This model is very well studied as it is the simplest model which involves tunneling through a barrier, so it provides a conceptual understanding to tunneling along a reaction coordinate.
The potential is visualized below along with the two lowest energy solutions:
Note, however, that in order to make a good comparison to the particle in a box, one must think about two times the particle in a box energy. This is because the double-well potential is really two infinite square-wells with a finite barrier between them. So, this tells you that in the limit that the barrier in the middle becomes infinitely large, there will be two ground state solutions with equal energy. Namely, the particle could be on the left or on the right.
So, as we lower the barrier and the particle is allowed to tunnel between the two wells, this degeneracy will be broken. These are the two wavefunctions which you see in the diagrams above. Namely, the solutions very closely resemble a symmetric and an antisymmetric linear combination of the ground state of a particle in an infinite box.
So, this has the effect of splitting the degeneracy into two states, one of which is higher in energy and one of which is lower in energy than particle in a box.
Why should the ground state of this double well be lower in energy than the ground state of a particle in a box where the particle stays on only one side of the barrier?
One way to think about this is that tunneling allows the particle to "leak" into the other box, which has the effect of making the effective box size of the particle larger.
This is all very qualitative, but the details of the math aren't really all that much more enlightening than simply looking at the qualitative features of the problem. The most important point I think is to recognize that tunneling through this barrier has the effect of splitting an otherwise degenerate state into a lower and higher energy state.