Calculating area of visible sky
Approximating the Earth as a sphere with radius $R$, then when viewing from a height $h$ above the surface, the Earth blocks out a cone of some opening angle $2\vartheta$, where $\csc\vartheta = 1+\frac{h}{R}$. Thus, the visible portion has a solid angle of
$$\Omega = 2\pi\left(1+\cos\vartheta\right) = 2\pi\left(1+\frac{\sqrt{h^2+2Rh}}{R+h}\right)$$
steradians. Divide this by $4\pi$ to obtain the fractional area of the the visible sky, compared to what you could have if the Earth wasn't blocking your view, since a full sphere subtends a solid angle of $4\pi$ steradians. That is probably a more natural measure of the visible sky than a literal area.

For an actual area, you need some sort of reference distance $r$ to measure from, with the visible sky a distance $r$ away having area $A = \Omega r^2$.