I wouldn't use Van Kampen. I would just note that this space deformation retracts to the union of the two diagonals of the square which is an X shape with the four ends glued together. This in turn is homotopy equivalent to a wedge of three circles by contracting one of the edges. Do you know the fundamental group of this?
Or to use Van Kampen, a neighborhood of the perimeter has free fundamental group generated by the four edges. Gluing in the interior of the square adds the relation that the product of these four generators is trivial. So we have $\langle a,b,c,d\,|\, abcd=1\rangle$, which isomorphic to a free group on three generators.