Solve: First, consider parallelograms whose sides are not parallel to side $BC$. Extend the sides of these parallelograms, and they will intersect with side $BC$. Therefore, these parallelograms correspond to four points or three points on side $BC$ (one vertex is on $BC$). Extend sides $AB$ and $AC$ by $\frac{1}{n}$, to get $\triangle AB'C'$. Thus, the parallelograms inside $\triangle ABC$ with sides not parallel to $BC$ correspond to four points on $B'C'$.

Conversely, for any four points on $B'C'$, draw lines parallel to $AB'$ through the first two points (closer to $B'$) and lines parallel to $AC'$ through the last two points (closer to $C'$). These lines intersect to form a unique parallelogram.

Thus, we can establish a correspondence: any parallelogram with sides not parallel to $BC$ corresponds to four points on $B'C'$, and the correspondence rule is:
in the set $\{1,2, \cdots, n+2\}$, a four-element ordered tuple.

The number of parallelograms is $C_{n+2}^{4}$.
Therefore, the total number of parallelograms is $3 C_{n+2}^{4}$.