The $n$-dimensional unit sphere (in the indexing in which the Earth is a 3-dimensional sphere) has an inner volume which is its surface area divided by $n$ (by considering each tiny patch of its surface as the base of a figure tapering down to its center), and at the same time, by the argument that projecting two dimensions outwards to the enveloping "cylinder" is area-preserving (stretching latitudinally and longitudinally in precise cancellation, as in the Lambert map projection), we find that the unit $n$-sphere's surface area is $\tau$ times the unit $(n - 2)$-sphere's volume.

[Footnote: Here, whenever I say “(inner) volume”, I mean in the sense appropriate to the number of dimensions, and similarly for “surface area”; thus, for example the “surface area” of a circle is its perimeter, and the “volume” of a circle is its area]

Putting these last two facts together, the unit $n$-sphere's volume is $\frac{\tau}{n}$ times the unit $(n - 2)$-sphere's volume ($\tau$ here being the circumference of a unit circle).

Letting $V(n)$ be the unit $n$-sphere's volume, and letting $D(n) = \frac{V(n + 1)}{V(n)}$, we get that $\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times \ldots \times \frac{n}{n + 1}$ comes out to $\frac{V(n + 1)/V(1)}{V(n) / V(0)} = \frac{V(0)}{V(1)} D(n) = \frac{1}{2} D(n)$. And of course $\frac{2}{1} \times \frac{4}{3} \times \frac{6}{5} \times \ldots$, with as many factors, is just this same thing times $n + 1$. Putting these together, the Wallis product up through $\frac{n}{n + 1} \times \frac{n}{n - 1}$ is $\frac{n + 1}{4} D(n)^2$.

But considering the $(n + 1)$-sphere's volume in terms of its cross-sectional $n$-spheres, we also have that $D(n)$ is twice the average value of $f(x)^n$ as $x$ runs from $0$ to $1$, where $f(x) = \sqrt{1 - x^2}$ is the cross-sectional radius of at height $x$ above the center of a unit sphere. Thus $D(n)$ is a decreasing function of $n$. And our recurrence $V(n) = \frac{\tau}{n} \times V(n - 2)$ is equivalently the recurrence $D(n - 2) \times D(n - 1) = \frac{\tau}{n}$. As a monotonic sequence satisfying this recurrence, we get that $D(n)^2 \sim \frac{\tau}{n}$ as $n$ goes to infinity (since $D(n)^2$ is inbetween $D(n - 1) \times D(n)$ and $D(n) \times D(n + 1)$, and our recurrence tells us each of these is $\sim \frac{\tau}{n}$).

Combining the last two paragraphs, we get that the Wallis product in toto comes out to $\frac{\tau}{4}$ (i.e., $\frac{\pi}{2}$). Ta-da!

This is actually related to Wallis's original proof (which was done not in terms of spheres but in terms of some messy integrals; even messier perhaps by virtue of being worked out before the general integral calculus had been developed!), but that is as this seen through a glass very darkly. I would be very curious to see if anyone has written on seeing the Wallis product this way in terms of sphere volumes before.