How to evaluate $\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx$?
Here is my trial, which is partially successful but still not fully answering to your question.
Using some coordinate change, I derived that
$$ I_{n} := \int_{0}^{\infty} \mathrm{erfc}^{n}(x) \, dx = \frac{1}{\sqrt{n}} \left( \frac{2}{\sqrt{\pi}} \right)^{n-1} \int_{T^{n-1}} \int_{0}^{\infty} s^{n-1}e^{-(|x|^{2}-1)s^{2}} \mathrm{erfc}(s) \, ds d\sigma_{x}, \tag{*} $$
where $T^{n-1} = \{ x \in [0, \infty)^{n} : x_{1} + \cdots + x_{n} = \sqrt{n} \}$ is an $(n-1)$-simplex and $d\sigma_{x}$ is the surface measure on $T^{n-1}$.
Case $n = 1$. Since $T^{0} = \{1\}$ and $d\sigma_{x} = \delta(x-1)$ is a unit mass located at $x = 1$, the equation $\text{(*)}$ gives no new information.
Case $n = 2.$ It is easy to find that
$$ \int_{0}^{\infty} s e^{-(|x|^{2}-1)s^{2}} \mathrm{erfc}(s) \, ds = \frac{1}{2|x|(|x|+1)}$$
and by suitable parametrization of the line segment $T^{1}$, we get
\begin{align*} I_{2} &= \sqrt{\frac{2}{\pi}} \int_{T^{1}} \frac{1}{2|x|(|x|+1)} \, d\sigma_{x} \\ &= \sqrt{\frac{2}{\pi}} \int_{0}^{\sqrt{2}} \frac{\sqrt{2} \, dt}{2\sqrt{t^{2} + (\sqrt{2}-t)^{2}} ( \sqrt{t^{2} + (\sqrt{2}-t)^{2}} + 1)} \\ &= \sqrt{\frac{2}{\pi}} \int_{0}^{\frac{\pi}{4}} \frac{d\theta}{1+\cos\theta} = \sqrt{\frac{2}{\pi}} \tan\left(\frac{\pi}{8}\right) = \frac{2 - \sqrt{2}}{\sqrt{\pi}}. \end{align*}
We can also write
$$ I_{2} = \boxed{ \displaystyle \frac{2}{\sqrt{\pi}} - \frac{2\sqrt{2}}{\pi^{3/2}} \arctan(\infty) }. $$
Case $n = 3$. We have
$$ \int_{0}^{\infty} s^{2}e^{-(|x|^{2}-1)s^{2}} \mathrm{erfc}(s) \, ds = \frac{1}{2\sqrt{\pi}(|x|^{2}-1)} \left( \frac{\arctan\sqrt{|x|^{2}-1}}{\sqrt{|x|^{2}-1}} - \frac{1}{|x|^{2}} \right). $$
Now we know that $T^{2}$ is a regular triangle with side length $\sqrt{6}$. Thus by introducing polar coordinate change, we get
\begin{align*} I_{3} &= \frac{2}{\pi^{3/2}\sqrt{3}} \int_{T^{2}} \frac{1}{|x|^{2}-1} \left( \frac{\arctan\sqrt{|x|^{2}-1}}{\sqrt{|x|^{2}-1}} - \frac{1}{|x|^{2}} \right) \, d\sigma_{x} \\ &= \frac{4\sqrt{3}}{\pi^{3/2}} \int_{0}^{\frac{\pi}{3}} \int_{0}^{\frac{\sec\theta}{\sqrt{2}}} \frac{1}{r} \left( \frac{\arctan r}{r} - \frac{1}{r^{2} + 1} \right) \, drd\theta \\ &= \frac{4\sqrt{3}}{\pi^{3/2}} \int_{0}^{\frac{\pi}{3}} \left( 1- \frac{\arctan\left(\sec\theta / \sqrt{2}\right)}{\sec\theta / \sqrt{2}} \right) \, d\theta \\ &= \frac{4\sqrt{3}}{\pi^{3/2}} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \left( 1 - \frac{\arctan u}{u} \right) \frac{du}{u\sqrt{2u^{2} - 1}}. \end{align*}
I haven't tried the last integral, but Mathematica suggests that
$$ \int_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \left( 1 - \frac{\arctan u}{u} \right) \frac{du}{u\sqrt{2u^{2} - 1}} = \frac{\sqrt{3}\pi}{4} -\sqrt{\frac{3}{2}} \arctan\left( \sqrt{2} \right).$$
This gives
$$ I_{3} = \boxed{ \displaystyle \frac{3}{\sqrt{\pi}} - \frac{6\sqrt{2}}{\pi^{3/2}} \arctan\left( \sqrt{2} \right) }, $$
which can be numerically checked.
I am trying to generalize this calculation for higher dimensions, but it seems somewhat daunting. For example, if $n = 4$ we have to evaluate
$$ \int_{0}^{\infty} \mathrm{erfc}^{4}(x) \, dx = \frac{1}{\pi^{3/2}} \int_{T^{3}} \frac{2|x|+1}{(|x|+1)^{2}|x|^{3}} \, d\sigma_{x}. $$
Nevertheless, using the formula
$$ \int_{T^{n-1}} f(|x|) \, d\sigma_{x} = n! \int_{0}^{\sqrt{\frac{1}{n-1}}} \int_{0}^{\sqrt{\frac{n}{n-2}}s_{1}} \cdots \int_{0}^{\sqrt{\frac{3}{1}}s_{n-2}} f\left( \sqrt{1+s_{1}^{2} + \cdots + s_{n-1}^{2}} \right) \, ds_{n-1} \cdots ds_{1} $$
together with aid of Mathematica, I was able to evaluate $I_{4}$ and it was
$$ I_{4} = \boxed{ \displaystyle \frac{4}{\sqrt{\pi}} - \frac{24\sqrt{2}}{\pi^{3/2}} \arctan \left( \frac{1}{\sqrt{8}} \right) }. $$
These series of observations lead us to believe that $I_{n}$ is of the form
$$ I_{n} = \frac{n}{\sqrt{\pi}} - \frac{n!\sqrt{2}}{\pi^{3/2}} \arctan \alpha_{n} $$
for some reasonably nice $\alpha_{n}$ (with $\alpha_{1} = 0$, $\alpha_{2} = \infty$, $\alpha_{3} = \sqrt{2}$ and $\alpha_{4} = \frac{1}{\sqrt{8}}$), but inverse symbolic calculations show that this seems no longer the case for $n \geq 5$.
Indeed, for $n = 5$ we have
$$ I_{5} = \frac{5}{\sqrt{\pi}} - \frac{80\sqrt{2}}{\pi^{3/2}} \arctan \sqrt{\frac{2}{3}} + \frac{240\sqrt{2}}{\pi^{5/2}} A\left( \sqrt{\frac{5}{3}}, \sqrt{\frac{3}{2}}, \sqrt{\frac{4}{5}} \right), $$
where $A(p, q, r)$ is the generalized Ahmed's integral. I have no idea whether it will reduce to a familiar closed form or not.
Integrating by parts twice, we get
$$ \begin{align} \int_{0}^{\infty} \text{erfc}^{3}(x) \, dx &= x \, \text{erfc}^{3}(x) \Bigg|^{\infty}_{0} + \frac{6} {\sqrt{\pi}} \int_{0}^{\infty}x \, \text{erfc}^{2}(x) e^{-x^{2}} \, dx \\ &= \frac{6} {\sqrt{\pi}} \int_{0}^{\infty}x \, \text{erfc}^{2}(x) e^{-x^{2}} \, dx \\ &= - \frac{3 \, \text{erfc}(x) e^{-x^{2}}}{\sqrt{\pi}} \Bigg|_{0}^{\infty} - \frac{12}{\pi} \int_{0}^{\infty}\text{erfc}(x) e^{-2x^{2}} \, dx \\ &= \frac{3}{\sqrt{\pi}} - \frac{12}{\pi} \int_{0}^{\infty} \text{erfc}(x) e^{-2x^{2}} \, dx. \end{align}$$
And using the integral definition of the complementary error function, we get
$$ \begin{align} \int_{0}^{\infty} \text{erfc}(x) e^{-2x^{2}} \, dx &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \int_{1}^{\infty} x e^{-x^{2}t^{2}} e^{-2x^{2}} \, dt \, dx \\ &= \frac{2}{\sqrt{\pi}} \int_{1}^{\infty} \int_{0}^{\infty} x e^{-(2+t^{2})x^{2}} \, dx \, dt \\ &= \frac{1}{\sqrt{\pi}} \int_{1}^{\infty} \frac{1}{2+t^{2}} \, dt \\ &= \frac{\sqrt{2}}{2 \sqrt{\pi}} \int_{1 / \sqrt{2}}^{\infty} \frac{1}{1+u^{2}} \, du \\ &= \frac{\sqrt{2}}{2 \sqrt{\pi}} \left[ \frac{\pi}{2} - \arctan \left(\frac{1}{\sqrt{2}} \right) \right] \\ &= \frac{\sqrt{2}}{2 \sqrt{\pi}} \, \arctan (\sqrt{2}). \end{align}$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \text{erfc}^{3}(x) \ dx &= \frac{3}{\sqrt{\pi}} - \frac{12}{\pi} \left( \frac{\sqrt{2}}{2 \sqrt{\pi}} \arctan \sqrt{2}\right) \\ &= \frac{3}{\sqrt{\pi}} - \frac{6 \sqrt{2}}{\pi^{3/2}}\, \arctan (\sqrt{2}). \end{align}$$
EDIT:
Again integrating by parts twice, we get
$$\int_{0}^{\infty} \text{erfc}^{4}(x) \ dx = \frac{4}{\sqrt{\pi}} + \frac{24}{\pi} \int_{0}^{\infty} \text{erfc}^{2}(x) e^{-2x^{2}} \, dx.$$
In general, for positive parameters $a,b$, and $p$,
$$ \begin{align} I(a,b,p) &= \int_{0}^{\infty} \text{erfc}(ax) \, \text{erfc}(bx) e^{-px^{2}} \, dx \\ &= \frac{4}{\pi} \int_{0}^{\infty} \int_{a}^{\infty} \int_{b}^{\infty} x^{2} e^{-x^{2}y^{2}} e^{-x^{2} z^{2}} e^{-px^{2}} \, dy \, dz \, dx \\ &= \frac{4}{\pi} \int_{a}^{\infty} \int_{b}^{\infty} \int_{0}^{\infty} x^{2} e^{-(p+y^{2}+z^{2})x^{2}} \, dx \, dy \, dz \\ &= \frac{1}{\sqrt{\pi}} \int_{a}^{\infty} \int_{b}^{\infty} \frac{1}{(p+y^{2}+z^{2})^{3/2}} \, dy \, dz. \end{align}$$
Let $y = \sqrt{p+z^{2}} \tan \theta$.
$$ \begin{align} I(a,b,p) &= \frac{1}{\sqrt{\pi}} \int_{a}^{\infty} \int_{\arctan(b / \sqrt{p+z^{2}})}^{\pi /2} \frac{\cos \theta}{p+z^{2}} \, d \theta \, dz \\ &= \frac{1}{\sqrt{\pi}} \int_{a}^{\infty} \frac{1}{p+z^{2}} \left(1- \frac{b}{\sqrt{p+b^{2}+z^{2}}} \right) \, dz \\ &= \frac{1}{\sqrt{\pi p}} \arctan \left(\frac{\sqrt{p}}{a} \right) - \frac{b}{\sqrt{\pi}} \int_{a}^{\infty} \frac{1}{(p+z^{2})\sqrt{p+b^{2}+z^{2}}} \, dz \end{align}$$
Now let $ \displaystyle t = \frac{1}{z}$.
$$ I(a,b,p) =\frac{1}{\sqrt{\pi p}} \arctan \left(\frac{\sqrt{p}}{a} \right) - \frac{b}{\sqrt{\pi}} \int_{0}^{1/a} \frac{t}{(1+pt^{2})\sqrt{1+b^{2}t^{2}+pt^{2}}} \, dt$$
Finally let $u^{2} = 1+b^{2}t^{2} + p t^{2}$.
$$ \begin{align} I(a,b,p) &= \frac{1}{\sqrt{\pi p}} \arctan \left(\frac{\sqrt{p}}{a} \right) - \frac{b}{\sqrt{\pi}} \int_{1}^{\sqrt{p+a^{2}+b^{2}}/a} \frac{1}{b^{2}+pu^{2}} \ du \\ &=\frac{1}{\sqrt{\pi p}} \arctan \left(\frac{\sqrt{p}}{a} \right) - \frac{1}{\sqrt{\pi p}}\int_{\sqrt{p} / b}^{\sqrt{p(p+a^{2}+b^{2)}}/(ab)} \frac{1}{1+w^{2}} \ dw \\ &= \frac{1}{\sqrt{\pi p}} \left[\arctan \left(\frac{\sqrt{p}}{a} \right) +\arctan \left(\frac{\sqrt{p}}{b} \right) - \arctan \left( \frac{\sqrt{p(p+a^{2}+b^{2})}}{ab}\right) \right] \end{align}$$
Therefore,
$$ I(1,1,2) = \int_{0}^{\infty} \text{erfc}^{2}(x) e^{-2x^{2}} \ dx = \frac{1}{\sqrt{2 \pi}} \left( 2 \arctan (\sqrt{2})- \arctan (2\sqrt{2}) \right) ,$$
and
$$ \begin{align} \int_{0}^{\infty} \text{erfc}^{4}(x) \ dx &= \frac{4}{\sqrt{\pi}} - \frac{12 \sqrt{2}}{\pi^{3/2}} \Big(2 \arctan (\sqrt{2})- \arctan (2 \sqrt{2}) \Big) \\ &= \frac{4}{\sqrt{\pi}} - \frac{24 \sqrt{2}}{\pi^{3/2}} \arctan \left( \frac{1}{2\sqrt{2}} \right). \end{align}$$
$I_5$ can be reduced to an integral of elementary functions. Namely,
\begin{eqnarray} I_5&=&\int_0^{\infty}\operatorname{erfc}^5x\,dx =\\ &=&\frac{5}{\pi^{1/2}}-\frac{240\sqrt{2}\arctan{\sqrt{2}}}{\pi^{5/2}}\left(\arctan2- \frac{\pi}{4}\right)+\frac{480}{\pi^{5/2}}\int_{\operatorname{arcsinh 1}}^{\infty}\frac{\arctan(\cosh x)}{3\cosh^2x-1}dx. \end{eqnarray}
Making the change of variables $z=e^{-x}$ in the remaining integral, one can rewrite it as \begin{eqnarray} \int_{\operatorname{arcsinh 1}}^{\infty}\frac{\arctan(\cosh x)}{3\cosh^2x-1}dx=\\ =\frac{1}{\sqrt3}\Im \int_{1+\sqrt2}^{\infty}\left(\frac{\log\left(1+i\frac{z+z^{-1}}{2}\right)}{z^2-\frac{2}{\sqrt3}z+1}-\frac{\log\left(1+i\frac{z+z^{-1}}{2}\right)}{z^2+\frac{2}{\sqrt3}z+1}\right)dz. \end{eqnarray} It is rather obvious that this can be written in terms of dilogarithm values (as I suggested in the previous version of my post). In particular, Mathematica can evaluate the integrals but I am too lazy to retype its long answer - see comment of Vladimir Reshetnikov below.
In order to obtain the first formula:
Integrate twice by parts to express $I_5$ in terms of $Q=\int_0^{\infty}e^{-2x^2}\operatorname{erfc}^3x\,dx$.
Then consider one-parameter deformation $Q(\alpha)=\int_0^{\infty}e^{-2x^2}\operatorname{erfc}^3(\alpha x)\,dx$.
The derivative $Q'(\alpha)$ is easily computable. Integrating it back and using that $Q(\infty)=0$, we get \begin{eqnarray} I_5=\frac{5}{\pi^{1/2}}-\frac{240\sqrt{2}\arctan{\sqrt{2}}}{\pi^{5/2}}\left(\arctan2- \frac{\pi}{4}\right)+\frac{480}{\pi^{5/2}}\int_1^{\infty}\frac{\arctan\sqrt{1+\alpha^2}}{\left(3\alpha^2+2\right)\sqrt{1+\alpha^2}}d\alpha. \end{eqnarray}
The remaining integral gives the above expression after the change of variables $\alpha =\sinh x$.