There are three possible shapes of parallelograms in this diagram which are shown below.

Each type has two adjacent sides paralleling to the two edges of the big equilateral triangle. Therefore, when the whole picture rotates $\pm 60^{\circ}$, one type of parallelograms will become another type. This means that the three sets of parallelograms are rotationally symmetric. As a result, it is sufficient to just count one type of parallelograms. The final result is three times of that number.

Without loss of generality, let's count the middle type of parallelograms above. For this, we can extend the bottom of this triangle by one row as shown below.

Extending the four edges of a parallelogram to meet the newly added bottom line will lead to four distinct meeting points. Conversely, any four distinct points on the bottom edge will map to a qualified parallelogram. This means that there is a bijection between the set of all such parallelograms and the set of collections of $4$ distinct points.

Because there are $\binom{n+1}{4}$ sets containing $4$ different points on the bottom line, therefore the final result is $$\boxed{3\binom{n+1}{4}}$$