Denote by $H$ the height of the pyramid, by $R$ the radius of the circumcircle of the base, by $h$ the height of the triangle $SA_1A_5$, by $a$ the segment $A_1A_5$, by $A$ the midpoint of $A_1A_5$, by $O$ the center of the base and by $S_{base}$ the are of the base.
The volume $V = \frac{1}{3}*S_{base}*H$ and $S=\frac{ah}{2}$
From the triangle $SOA$ we get $H=h*sin \alpha $ and $AO=\frac{R}{2}=h*cos\alpha$. Then $S_{base}=\frac{(2*h*cos\alpha)^2*sin\frac{\pi}{3}}{2}=(h*cos\alpha)^2$
From the triangle $A_1AO$ we get $AO=\frac{R}{2}=\frac{AA_1}{2*\sqrt{3}}=\frac{a}{4*\sqrt{3}}=h*cos\alpha$, hence $h=\frac{a}{4*\sqrt{3}*cos\alpha}=\frac{\frac{2S}{h}}{4*\sqrt{3}*cos\alpha}$ $\Leftrightarrow$ $h=\sqrt{\frac{S}{2*\sqrt{3}*cos\alpha}}$
From the above results $V=\frac{1}{3}*(h*cos\alpha)^2*h*sin \alpha=\frac{1}{3}*h^3*cos^2\alpha*sin\alpha$ $=\frac{S\sqrt{S}*\sqrt{cos\alpha}*sin\alpha}{2\sqrt{2}*\sqrt[4]{27}}$