Let $r_n$ denote the inradius of a regular $n$-simplex $\triangle^n$ in $\mathbb{R}^n$, and $a$ denote the uniform edge length.
It is well-known that
$r_1 = a \frac{1}{2} \\ r_2 = a \frac{1}{6} \sqrt{3} \\ r_3 = a \frac{1}{12} \sqrt{6}$
But how to generalize $r_n$ to arbitrary dimensions?
As we see in Height of $n$-simplex, the altitude (height) of a regular $n$-simplex of edge length $a$ is
$$h_n = a\sqrt{\frac{n+1}{2n}}.$$
The inradius $r_n$ is the distance from the centroid of the regular $n$-simplex to the center of a hyperface, so $r_n = \frac 1{n+1} h_n.$ Therefore
$$r_n = \frac a{n+1} \sqrt{\frac{n+1}{2n}} = \frac{a}{\sqrt{2n(n+1)}}.$$
You might check the proof of
Cartesian coordinates for vertices of a regular 16-simplex?
which gives, via an inductive midpoint construction, the inradii as $h_n$ for a unit length regular $n$-simplex.
