First I'm going to declare some variables to be used in this solution: B is the area of triangle A1A3A5; C is the area of triangle A1A2A3, and x is the length of A1A3.

Then some setup.  Let O be the center of the dodecagon and M be the midpoint of A1A5.

Draw A1A3, A3A5, A5A7, A7A9, A9A11, A11A1.  This creates six small triangles hugging the perimeter of the dodecagon.  Each one has area of C.

Draw A1A5, A5A9, A9A1, A1O, A5O, A9O.  This creates six large triangles in the interior of the dodecagon.  Each one has area of B

The area of the dodecagon then equals 6B+6C.

Draw OA3.  OA3 and A1A5 are orthogonal and bisect each other at M.

The plane of triangle MOP is orthogonal to both the plane of the dodecagon and the plane of triangle PA1A5.  Then the measure of angle OMP is alpha.

Now, I'll solve a simplified version of the problem: "express the volume of the hexagonal pyramid PA1A3A5A7A9A11 in terms of alpha and area of A1A3A5".

The area of A1A3A5 is defined to be B.  But A1A3A5 is also a 30-30-120 triangle and has A1A3=A3A5=x as two of its sides.  Also A3M (congruent to MO) is the altitude perpendicular to A1A5.  Some elementary trig then tells us that B = z^2*sqrt(3)/4 and MO=A3M = sqrt(B/sqrt(3)).

Then altitude PO of the pyramid can be found from tan(alpha) = PO/OM which implies PO = tan(alpha)*sqrt(B/sqrt(3))

So the volume of the hexagon pyramid can be written as (1/3)*(6B)*(tan(alpha)*sqrt(B/sqrt(3))).

Now lets head back to the dodecagon pyramid.

First, lets add the small triangles back in.  Then the volume can be written as (1/3)*(6B+6C)*(tan(alpha)*sqrt(B/sqrt(3))).

Now lets get area C in terms of B.  A1A2A3 has area C, is a 15-15-150 triangle, and has x=A1A3 as its long side.  Some more elementary trig (and knowing tan(15)=2-sqrt(3)) tells us that C=x^2*(2-sqrt(3))/4.

Then the ratio C/B = (2-sqrt(3))/sqrt(3).  Then C=B*(2-sqrt(3))/sqrt(3).

Plug that into our expression for the volume and simplify:

(1/3)*(6B+6B*(2-sqrt(3))/sqrt(3))*(tan(alpha)*sqrt(B/sqrt(3)))

= [2 + 2*(2-sqrt(3))/sqrt(3)] / quadrt(3) * tan(alpha) * B^(3/2)

= (4 / 3^(3/4)) * tan(alpha) * B^(3/2)

But we still have one more task, to get area B in terms of given area S.  Triangle OA1A5 is the orthogonal projection of PA1A5 onto the plane of the dodecagon. Since they share side A1A5, the thing that changes is altitude MP projected onto altitude MO.  

MO/MP = cos(alpha), so then B/S = cos(alpha) implies B = S*cos(alpha).

Then one last substitution gives us the volume of the pyramid as

(4 / 3^(3/4)) * tan(alpha) * cos^(3/2)(alpha) * S^(3/2)

= (4 / 3^(3/4)) * sin(alpha) * sqrt(cos(alpha)) * S^(3/2)