The Wallis sieve is a variation on Sierpinski's carpet, where you start with a square, and in the $i$th step you divide each square into $(2i+1)^2$ smaller squares and remove the middle one. The total area of the resulting fractal is equal to $\pi/4$ by the Wallis product.
I define the $n$-dimensional Wallis sieve as the fractal generated as follows: Start with an $n$-cube of side length $2$. In the $i$th step, divide the $n$-cube into $3^n$ smaller $n$-cuboids by dividing each dimension in a $i:1:i$ ratio. Remove every $n$-cuboid that is "in the middle" of its parent $n$-cube with respect to at least $2$ of the $n$ dimensions. Finally, divide each $n$-cuboid into smaller $n$-cubes, as if the original $n$-cube had been divided into $(2i+1)^n$ smaller $n$-cubes.
In the $i$th step, this operation reduces the total hypervolume by $$\frac{(2i+n)(2i)^{n-1}}{(2i+1)^n}$$
Proof: When going from dimension $n$ to $n+1$, we have stretched copies of the original $n$-cube in $i:1:i$ ratios along the new dimension. If there were originally $p$ $n$-cuboids that are not "in the middle" of its parent $n$-cube with respect to any dimension and $q$ of those that are with respect to exactly $1$ dimension, now there will be $p$ and $q$ respectively in each outer segment and $0$ and $p$ respectively in the middle segment, $2p$ and $2q+p$ respectively in total. Solving the recurrence relation with starting conditions $p_0=1$, $q_0=0$ for a single point we get $p_n=2^n$, $q_n=n2^{n-1}$. The hypervolume of $n$-cuboids that are not "in the middle" of its parent $n$-cube with respect to any dimension, as a fraction of the parent $n$-cube, is $\frac{i^n}{(2i+1)^n}$, and that of those that are with respect to exactly $1$ dimension is $\frac{i^{n-1}}{(2i+1)^n}$. Those are the only two types of $n$-cuboids that do not get removed, and summing their hypervolumes gives the above expression.
It follows that the total remaining hypervolume is equal to $$2^n\prod_{i=1}^{\infty}{\frac{(2i+n)(2i)^{n-1}}{(2i+1)^n}}$$
For $n=0$, we have a single point, which is also the unit $0$-ball.
For $n=1$, no line segments are ever removed, and we are left with the original line segment of length 2, which is also the unit $1$-ball.
For $n=2$, we have the Wallis sieve of side length $2$, with area equal to $\pi$, which is equal to the area of the unit circle.
For $n=3$, we have a variation on the Menger sponge of side length $2$. The volume of the resulting fractal is equal to $\frac{4}{3}\pi$, which is equal to the volume of the unit ball. Reference: https://community.wolfram.com/groups/-/m/t/822984.
I have calculated the results for small values of $n\ge4$ using Wolfram Mathematica, and they all agree with the hypervolume of the corresponding unit $n$-ball, however I don't know how to prove it.
Substituting in formulas for the volume of the $n$-ball, one would need to prove either $$2^n\prod_{i=1}^{\infty}{\frac{(2i+n)(2i)^{n-1}}{(2i+1)^n}}=\frac {\pi ^{n/2}}{\operatorname {\Gamma } \left({\frac {n}{2}}+1\right)}$$ or $$\prod_{i=1}^{\infty}{\frac{(2i+n)(2i)^{n-1}}{(2i+1)^n}}=\frac {(\pi /2)^{\left\lfloor {\frac {n}{2}}\right\rfloor }}{n!!}$$
Ideally I would like a geometric proof showing how the removed sections can be assembled to exactly form the difference between the $n$-cube and the $n$-ball, at least in 2 or 3 dimensions.