$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Bd}{\partial}\DeclareMathOperator{\vol}{vol}$The formulas are no accident, but not especially deep. The explanation comes down to a couple of geometric observations.

If $X$ is the closure of a bounded open set in the Euclidean space $\Reals^{n}$ (such as a solid ball, or a bounded polytope, or an ellipsoid) and if $a > 0$ is real, then the image $aX$ of $X$ under the mapping $x \mapsto ax$ (uniform scaling by a factor of $a$ about the origin) satisfies
$$
\vol_{n}(aX) = a^{n} \vol_{n}(X).
$$
More generally, if $X$ is a closed, bounded, piecewise-smooth $k$-dimensional manifold in $\Reals^{n}$, then scaling $X$ by a factor of $a$ multiplies the volume by $a^{k}$.

If $X \subset \Reals^{n}$ is a bounded, $n$-dimensional intersection of closed half-spaces whose boundaries lie at unit distance from the origin, then scaling $X$ by $a = (1 + h)$ "adds a shell of uniform thickness $h$ to $X$ (modulo behavior along intersections of hyperplanes)". The volume of this shell is equal to $h$ times the $(n - 1)$-dimensional measure of the boundary of $X$, up to added terms of higher order in $h$ (i.e., terms whose total contribution to the $n$-dimensional volume of the shell is negligible as $h \to 0$).

If $X$ satisfies Property 2. (e.g., $X$ is a ball or cube or simplex of "unit radius" centered at the origin), then
$$
h \vol_{n-1}(\Bd X) \approx \vol_{n}\bigl[(1 + h)X \setminus X\bigr],
$$
or
$$
\vol_{n-1}(\Bd X) \approx \frac{(1 + h)^{n} - 1}{h}\, \vol_{n}(X).
\tag{1}
$$
The approximation becomes exact in the limit as $h \to 0$:
$$
\vol_{n-1}(\Bd X)
= \lim_{h \to 0} \frac{(1 + h)^{n} - 1}{h}\, \vol_{n}(X)
= \frac{d}{dt}\bigg|_{t = 1} \vol_{n}(tX).
\tag{2}
$$
By Property 1., if $r > 0$, then
$$
\vol_{n-1}\bigl(\Bd (rX)\bigr)
= r^{n-1}\vol_{n-1}(\Bd X)
= \lim_{h \to 0} \frac{(1 + h)^{n}r^{n} - r^{n}}{rh}\, \vol_{n}(X)
= \frac{d}{dt}\bigg|_{t = r} \vol_{n}(tX).
\tag{3}
$$
In words, the $(n - 1)$-dimensional volume of $\Bd(rX)$ is the derivative with respect to $r$ of the $n$-dimensional volume of $rX$.

This argument fails for non-cubical boxes and ellipsoids (to name two) because for these objects, uniform scaling about an arbitrary point does not add a shell of uniform thickness (i.e., Property 2. fails). Equivalently, adding a shell of uniform thickness does not yield a new region similar to (i.e., obtained by uniform scaling from) the original.

(The argument also fails for cubes (etc.) not centered at the origin, again because "off-center" scaling does not add a shell of uniform thickness.)

In more detail:

Scaling a non-square rectangle adds "thicker area" to the pair of short sides than to the long pair. Equivalently, adding a shell of uniform thickness around a non-square rectangle yields a rectangle *having different proportions* than the original rectangle.

Scaling a non-circular ellipse adds thicker area near the ends of the major axis. Equivalently, adding a uniform shell around a non-circular ellipse yields a non-elliptical region. (The principle that "the derivative of area is length" fails drastically for ellipses: The area of an ellipse is proportional to the product of the axes, while the arc length is a *non-elementary function* of the axes.)