Finding the exact number of unimodular triangulations of a cube in higher dimension is, I would say, out of reach. In fact, I would say the number is doubly exponential in n (that is, exponential in the number of vertices).

For dimension 3 I can reproduce David's count of 72 in a different fashion:

Every unimodular triangulation of the 3-cube uses one (and only one) of the four diameters of the cube. So, it suffices to count the triangulations that use one particular diameter. For this do the following: if you project the cube along the direction of the diameter used, you see a hexagon. The vertices to which the diameter is joined in your triangulation must form a sub polygon of this hexagon with the center inside. There are 18 possibilities:

• the whole hexagon.
• the six pentagons that you obtain leaving out one vertex of the hexagon.
• the nine quadrilaterals that you get leaving out two non-consecutive vertices.
• the two triangles that you get using alternate vertices.

You need to check that all possibilities do indeed extend to triangulations of the cube (so far we have only described the "star" of the diameter used) and that no possibility extends in two ways. Once you check this (there are 5 combinatorially different cases to understand) you will have proved that there are 18 triangulations using that diameter, and 72 in total.

Finding the exact number of unimodular triangulations of a cube in higher dimension is, I would say, out of reach. In fact, I would say the number is doubly exponential in n (that is, exponential in the number of vertices). [Update: see my other answer for a doubly exponential lower bound]

For dimension 3 I can reproduce David's count of 72 in a different fashion:

Every unimodular triangulation of the 3-cube uses one (and only one) of the four diameters of the cube. So, it suffices to count the triangulations that use one particular diameter. For this do the following: if you project the cube along the direction of the diameter used, you see a hexagon. The vertices to which the diameter is joined in your triangulation must form a sub polygon of this hexagon with the center inside. There are 18 possibilities:

• the whole hexagon.
• the six pentagons that you obtain leaving out one vertex of the hexagon.
• the nine quadrilaterals that you get leaving out two non-consecutive vertices.
• the two triangles that you get using alternate vertices.

You need to check that all possibilities do indeed extend to triangulations of the cube (so far we have only described the "star" of the diameter used) and that no possibility extends in two ways. Once you check this (there are 5 combinatorially different cases to understand) you will have proved that there are 18 triangulations using that diameter, and 72 in total.