Part 1: Initial assumptions
If we were to assume that the elevator cable is visible from any distance (assuming no obstacles), then this is actually a simple exercise in geometry.
Here's a quick diagram I drew (note: elevator length not to scale):

The black circle is the planet, the blue line is the elevator, and the red lines are imaginary lines tangent to points on the planet's surface.
At any point on the planet's surface, we can draw a line tangent to it that will intersect the elevator cable at some point. We can then draw a line from that point to the center of the planet (this line will be perpendicular to the tangent line):

The computation for any length is simple. At some height $h$, we have
$$\text{Length of tangent line}=\sqrt{R_{\text{planet}}^2+(R_{\text{planet}}+h)^2}$$
We also have
$$\text{Angle between point on surface and horizontal}=\arccos\left(\frac{R_{\text{planet}}}{R_{\text{planet}}+h}\right)$$
You should be able to find the area from this, which I leave as an exercise for the reader. Hint: Use the arc length of the central cross-section of the planet.
This whole approach fails, however, if we consider that the Earth is not a perfect sphere, and is not smooth. A better approximation is to treat it as an oblate spheroid. We can describe it in Cartesian coordinates by the equation
$$\frac{x^2}{6,378,137^2}+\frac{y^2}{6,378,137^2}+\frac{z^2}{6,356,752^2}=1$$
Using spherical coordinates, we then have
$$\frac{r^2\sin^2\theta\cos^2\varphi}{6,378,137^2}+\frac{r^2\sin^2\theta\sin^2\varphi}{6,378,137^2}+\frac{r^2\cos^2\theta}{6,356,752^2}=1$$
Doing some algebra, we have
$$r=r(\theta,\varphi)=\left(\sqrt{\frac{\sin^2\theta\cos^2\varphi}{6,378,137^2}+\frac{\sin^2\theta\sin^2\varphi}{6,378,137^2}+\frac{\cos^2\theta}{6,356,752^2}}\right)^{-1}$$
We can then set some $(\theta_0,\varphi_0)$ to be the location of the elevator, and make the changes
$$\theta\to\theta-\theta_0,\qquad\varphi\to\varphi-\varphi_0$$
At this point, you can use a similar method to the one used in the spherical approximation part.
Part 2: Accounting for visibility problems
Here, we need to discuss angular diameter. The angular diameter of an object $d$ meters wide at a distance of $D$ meters is
$$\delta=2\arctan\left(\frac{d}{2D}\right)$$
In your case, $d=$10 mile$=s$16,000 meters. According to Wikipedia, the minimum angular resolution the human eye can see is about 60 arcseconds, or 1/60 of a degree. Plugging this in, we have
$$D=\frac{1}{2}\left(\frac{16,000}{\tan(1/120)}\right)$$
I find a distance of about 55,000 kilometers.
[Under construction]