= Solution
For a <conductor in electrostatic equilibrium>, the interior field is zero and the surface is equipotential. The tangential exterior field is zero. A thin Gaussian pillbox gives the normal jump $E_{\rm out}\cdot\mathbf n-E_{\rm in}\cdot\mathbf n=\sigma/\epsilon_0$. Thus, in vacuum and with $\mathbf n$ pointing out of the conductor,
$$
\boxed{\mathbf E_{\rm out}=\frac{\sigma}{\epsilon_0}\mathbf n.}
$$
These are the <electrostatic boundary conditions at a conductor>.
To obtain <electrostatic energy> in field form, integrate over the space outside the conductors, cutting it off first at a large sphere. From the <Maxwell equations>, $\nabla\cdot\mathbf E=\rho/\epsilon_0$ and $\mathbf E=-\nabla\phi$, so
$$
\nabla\cdot(\phi\mathbf E)=\frac{\rho\phi}{\epsilon_0}-|\mathbf E|^2.
$$
On a conductor boundary, the outward normal of the integration region points into the conductor, so its flux contribution is $-\phi_iQ_i/\epsilon_0$. At the remote outer boundary the flux integral tends to zero: for a localized charge distribution, $\phi=O(1/r)$ and $\mathbf E=O(1/r^2)$. The <divergence theorem> therefore gives
$$
-\frac1{\epsilon_0}\sum_i\phi_iQ_i
=\frac1{\epsilon_0}\int_V\rho\phi\,d\tau
-\int_{\text{outside conductors}}|\mathbf E|^2\,d\tau.
$$
Since the field vanishes inside conductors,
$$
\boxed{W=\frac{\epsilon_0}{2}\int_{\mathbb R^3}|\mathbf E|^2\,d\tau.}
$$
The field integral is over all space, although the charges occupy a finite volume. For a finite outer integration boundary there is additionally the term $(\epsilon_0/2)\oint\phi\mathbf E\cdot d\mathbf S$.
For the <electrostatic energy of alternating-charge concentric shells>, the total enclosed charge after shell $n$ is
$$
Q_{\le n}=q+2q\sum_{j=2}^n(-1)^{j+1}=(-1)^{n+1}q.
$$
By spherical symmetry and <Gauss's law>, the field is zero for $r<r_1$, and in every subsequent gap, including outside the last shell,
$$
|\mathbf E|=\frac{|q|}{4\pi\epsilon_0r^2}.
$$
Although its direction alternates, its squared magnitude is the same in every gap. Consequently
$$
\boxed{W=\frac{\epsilon_0}{2}\,4\pi
\int_{r_1}^{\infty}\frac{q^2}{16\pi^2\epsilon_0^2r^4}r^2\,dr
=\frac{q^2}{8\pi\epsilon_0r_1}.}
$$
The coefficient of $1/r_1$ is $q^2/(8\pi\epsilon_0)$, independent of the other radii and of the number of shells.
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