Evaluate $\iiint_{[0,1]^3}\frac{dx\,dy\,dz}{(1+x^2+y^2+z^2)^2}$

Here's an intuitive solution involving surfaces of hypercubes.

Let $$v := \iiint_{[0,1]^3}\frac{dxdydz}{(1+x^2+y^2+z^2)^2}$$

be the value you're looking for.

Suppose a point-shaped light source shines uniformly from the origin of 4D space with intensity of $1$ watt per cubic radian. Let $p$ be a parametrization of a cubic piece $P$ of hyperplane, specifically $p(x,y,z) = (1,x,y,z)$ for $x, y, z \in [0,1]$.

It is clear that no part of this piece of hyperplane is in shadow. If we compute the irradiance $I(x,y,z)$ of arriving light in watts per cubic unit, we find

$$I(x,y,z) = \frac{1}{(1+x^2+y^2+z^2)^2}.$$

If we integrate this irradiance over the surface of $P$, we find that the total power arriving at $P$ is $v$ watts per cubic unit.

Now consider the surface of a hypercube with corners at $(\pm 1, \pm 1, \pm 1, \pm 1)$. We know that $P$ is an octant of one of the eight cells of this surface, and every other octant receives the same irradiance, so we know that the total power arriving at the hypercube is $64v$. However, the hypercube captures every bit of light emitted by our light source, so we know that $64v$ equals $2\pi^2$, the total wattage of light emitted by our light source.

Therefore $v = \frac{2\pi^2}{64} = \frac{\pi^2}{32}$.

Let us denote the integral by $I$, $$I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{(1+x^2+y^2+z^2)^2}=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{\infty} t \exp[-(1+x^2+y^2+z^2)t] dx dy dz dt$$ Due to the symmetry of $I$ in $x,y,z$; we can write $$I=\int_{0}^{\infty} dt ~t e^{-t} \left ( \int_{0}^{1} e^{-tx^2} dx \right)^3=\frac{\pi \sqrt{\pi}}{8} \int_{0}^{\infty} t e^{-t} \mbox{erf}~^3(\sqrt{t}) dt= \frac{\pi^2}{8} \int_{0}^{1} v^3 dv=\frac{\pi^2}{32}. $$ Here $\mbox{erf}(x)=\frac{2}{\sqrt{\pi}} \int_{0} ^{x} e^{-t^2} dt$ and we have used $\mbox{erf}(\sqrt{t})=v$.

I also post the screen shot of from Mathematica,

Since for any $a>0$ we have $\frac{1}{a^2}=\int_{0}^{+\infty} w e^{-aw}\,dw$, by Fubini's theorem the original integral can be written as

$$ \int_{0}^{+\infty} w e^{-w}\left(\frac{\sqrt{\pi}\,\text{Erf}(\sqrt{w})}{2\sqrt{w}}\right)^3\,dw\stackrel{w\mapsto w^2}{=}\frac{\pi\sqrt{\pi}}{4}\int_{0}^{+\infty}e^{-w^2}\text{Erf}^3(w)\,dw $$ and the RHS is clearly $$ \frac{\pi\sqrt{\pi}}{4}\left[\frac{\sqrt{\pi}}{8}\,\text{Erf}^4(w)\right]_{0}^{+\infty} =\color{red}{\frac{\pi^2}{32}}.$$

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Interesting side-note: the same approach in dimension $2$ gives a relation between $\iint_{(0,1)^2}\frac{dx\,dy}{(1+x^2+y^2)^2}=\frac{1}{\sqrt{2}}\arctan\frac{1}{\sqrt{2}}$ and $\int_{0}^{+\infty}\left(1-\text{Erf}^3(w)\right)\,dw$.
In particular it proves that $$ \int_{0}^{+\infty}\left(1-\text{Erf}^3(x)\right)\,dx = \frac{6\sqrt{2}}{\pi\sqrt{\pi}}\,\arctan\frac{1}{\sqrt{2}} $$ which is not recognized by Mathematica (or at least by my version).