Why don't higher order multipole moments "stack" like the monopole and dipole moments do? Electric charge is notorious for needing only a relatively small number of electrons or protons to move to produce macroscopically visible effects. Similarly, electric and magnetic dipoles produce big effects in capacitors and ferromagnets. Nature is full of objects that have permanent quadrupole moments and higher, especially the building blocks of ionically bound crystals, like table salt. Why don't these higher order $2^\ell$-pole moments seem to "stack up" in the same way the lower order moments do?
 A: This question, as with so many others in physics, can be answered by a simple scaling argument. Every permanent $2^\ell$-pole will have a characteristic size to the volume it takes up, call that $a$. The number of $2^\ell$-poles that can potentially stack up constructively is given by a simple ratio of volumes, say
$$N = \frac{L^3}{a^3},$$
in an $L\times L\times L$ cube. 
The multipole moment of the building block will have a normalization factor that can be arranged to be
$$Q_\ell^m \propto qa^\ell,$$
for some characteristic charge $q$ (usually around the electron charge). If we assume that we're dealing with the lowest non-vanishing multipole, so that it is translation invariant, then the total $2^\ell$-pole moment will be just the number of $2^\ell$-poles times the moment per $2^\ell$-pole, to get
\begin{align}
     Q_{\ell,\mathrm{tot}}^m &\propto Nqa^\ell \\
    & = q L^3 a^{\ell-3} \\
    & = q \left(\frac{a}{L}\right)^{\ell - 3} L^\ell = q' L^\ell.
\end{align}
The significance of the last line is that it is the same as exhcanging all of the microscopic $2^\ell$-poles for a single equivalent macroscopic one built with monopoles of characteristic charge $q'$ that get the same $2^\ell$-pole moment. For $\ell < 3$, size of the charges needed to build the equivalent macroscopic $2^\ell$-pole grows with $L$. For $\ell=3$, the octupole moment, the characteristic charge is constant (e.g. the net octupole moment of a salt cube is, up to a numerical factor of a few, the same as having a single uncompensated electron/proton on the corners of the cube). For $\ell > 3$ the charge needed to build the macroscopic equivalent $2^\ell$-pole is actually shrinks with growing $L$.
Granted, for fixed $q$ and $a$, the net $2^\ell$-pole moment always grows $\propto L^3$, but this change-out for an equivalent macroscopic $2^\ell$-pole places an abstract concept into concrete terms that make it easier to picture just how small the electric fields being produced are.
Another way to look at the issue is to consider how big the field produced at about the surface of the object is. For a $2^\ell$-pole the leading order term in the electric or magnetic field is proportional to $r^{-2-\ell}$. Since the the minimum observation distance is $r\sim L$, and the $2^\ell$-pole moment grows like $L^3$, the electric field at the minimum distance from the point-like equivalent $2^\ell$-pole is $\propto L^{1-\ell}$. With this way of measuring the "impact" of trying to stack multipoles, only the monopole moments stack well, with the dipole moments being marginal.
And all of this assumes that the "stacking" can be done perfectly, ignoring thermodynamic effects that lead to magnetic domains and crystal grains.
