There is a very simple geometric explanation for the fact that the constant of proportionality is 1 for the sphere's radius and the cube's half-width. In fact, this relationship also lets you define a sensible notion of a "half-width" of an arbitrary $n$-dimensional shape.
Pick an arbitrary shape and a point $O$ inside it. Suppose you enlarge the shape by a factor $\alpha \ll 1$ keeping $O$ fixed. Each surface element with area $dA$ at a position $\vec r$ relative to $O$ gets extruded into an approximate prism shape with base area $dA$ and offset $\alpha \vec r$. The corresponding additional volume is $\alpha \vec r\cdot \vec n dA$, where $\vec n$ is the normal vector at the surface element.
Now the quantity $\vec r \cdot \vec n$, call it the projected distance, has a natural geometric interpretation. It is simply the distance between $O$ and the tangent plane at the surface element. (Observe that for a sphere with $O$ at the center, it is always equal to the radius, and for a cube with $O$ at the center, it is always equal to the distance from the center to any face.)
Let $\hat r = A^{-1} \int \vec r \cdot \vec n dA$ be the mean projected distance over the surface of the shape. Then the change in volume by a scaling of $\alpha$ is simply $\delta V = \alpha \int \vec r\cdot \vec n dA = \alpha \hat r A$. In other words, a change of $\alpha \hat r$ in $\hat r$ corresponds to a changed of $\alpha \hat r A$ in $V$. So if you use $\hat r$ as the measure of the size of a shape, you find that $dV/d\hat r = A$. And since $\hat r$ equals the radius of a sphere and the half-width of a cube, the observation in question follows. This also implies the distance-to-face measure for regular polytopes that user9325 mentioned, but generalizes to other polytopes and curved shapes. (I'm not completely happy with the definition of $\hat r$ because it's not obvious that it is independent of the choice of $O$. If someone can see a more natural definition, please let me know.)
Imagine a vertical cylinder enclosing the sphere, with height $2r$, radius $r$, and open ends. This cylinder has surface area $4\pi r^2$. The trick is to show that if you slice the cylinder and the sphere into infinitesimally thin horizontal rings, then at a given height, the surface area of the spherical ring equals the surface area of the cylindrical ring. Thus the total surface areas are equal.
Suppose the cylindrical ring has has height $\delta h$, and therefore area $2\pi r \times \delta h$. If the ring is at a height $r\sin\theta$ above the equator of the sphere, with $-\pi < \theta < \pi$, then the spherical ring has radius $r\cos\theta$, but its surface is at an angle $\theta$ from the vertical. So its area is $2\pi r \cos \theta \times \delta h/\cos \theta$, which is the same as the cylindrical ring.
This really needs some nice pictures, but I have no skill in that direction.
Best Answer
Yes, the mapping preserves area of any shape. You can convince yourself of this by taking by small patches on the sphere, between two constant latitude lines and two longitude lines, which I believe is what they did with the state of Colorado and the sate of Wyoming. Anyway, any nice enough shape is made up, to sufficient accuracy, by a large number of these curved rectangles, and these quite definitely are mapped in an area-preserving manner. This is the oldest example of a "symplectic" map.
I do not know that much about the history of this exact example, but I do know that a book of Archimedes called The Method was thought to be lost until about 1900, and translations are available. See EUDOXUS and METHOD and SPHERE_AND_CYLINDER finally MOOSE_AND_SQUIRREL
Alright, somebody at Wikipedia is not paying attention. The equal area MAP projection is due to Archimedes.