I will generalize a bit and calculate the number of parallelograms (2 sets of parallel sides), trapezoids (1 set of parallel sides), and trapeziums (no sets of parallel sides) for which the total number of quadrilaterals can be split into.

Drawing in all the chords of the regular polygon forms the set of possible sides and diagonals of the quadrilaterals.  Note that only for even n there exist diameters of the circumcircle as possible chords to pick from. 

First thing to note is that all the quadrilaterals are cyclic polygons whose circumcircle is the same as the polygon.  This has two important implications. First is all possible parallelograms are rectangles, whose diagonals are diameters.  Second is all possible trapezoids are isosceles trapezoids, which cannot have a diameter as a diagonal.

Let Q(n) be the total number of quadrilaterals that can be formed for a regular polygon of n sides. Similarly, let R(n) be the total number of rectangles, let S(n) be the total number of trapezoids, and let T(n) be the total number of trapeziums.

Since rectangles, trapezoids, and trapeziums represent the possible quadrilaterals then Q(n) = R(n) + S(n) + T(n).

Q(n) can be calculated simply as the number of ways to choose 4 vertices out of n.  Then Q(n) = nC4 = n(n-1)(n-2)(n-3)/24

R(n) can be calculated by choosing which two diameters are the diagonals of the rectangle.  Then for odd n: R(n) = 0 and for even n: R(n) = (n/2)C2 = (n/2)*(n/2-1)/2 = n(n-2)/8

S(n) can be calculated by counting the number of obtuse angles possible for the more clockwise of the trapezoid's two obtuse angles.  To do that take each vertex as the most clockwise of the three vertices defining the angle and count the number of ways to choose the other two points in the half-circle (excluding the possible point that would form a diameter) from that chosen vertex. 

For odd n there will be (n-1)/2 other vertices to choose from and for even n there will be (n-2)/2 other vertices.  Then for odd n: S(n) = n*((n-1)/2)C2 = n*((n-1)/2)*((n-1)/2-1)/2 = n(n-1)(n-3)/8 and for even n: S(n) = n*((n-2)/2)C2 = n*((n-2)/2)*((n-2)/2-1)/2 = n(n-2)(n-4)/8.

Evaluating T(n) then is a matter of substituting the odd and even forms of Q(n), R(n), S(n) into Q(n) = R(n) + S(n) + T(n).  Then for odd n: T(n) = n(n-1)(n-2)(n-3)/24 - n(n-1)(n-3)/8 = n(n-1)(n-3)(n-5)/24 and for even n: T(n) = n(n-1)(n-2)(n-3)/24 - n(n-2)/8 - n(n-2)(n-4)/8 = n(n-2)(n-3)(n-4)/24.

In summary the following table shows the number of a specific quadrilateral for odd/even n:

n-gon sides |      n is odd       |     n is even
------------+---------------------+---------------------
Total Quads | n(n-1)(n-2)(n-3)/24 | n(n-1)(n-2)(n-3)/24
Rectangles  |         0           |      n(n-2)/8
Trapezoids  |    n(n-1)(n-3)/8    |    n(n-2)(n-4)/8
Trapeziums  | n(n-1)(n-3)(n-5)/24 | n(n-2)(n-3)(n-4)/24