all multipolar terms of nuclear fields must be confined? When usually speaking about QCD confinement, one refers to the fact that color charges over lengths above the hadron size are fully neutral, which implies that monopolar charges over an extended region will be zero. But this doesn't exclude non-zero dipolar, or quadrupolar moments of color charge.
Are macroscopic fields of dipolar, or multipolar color charge moments excluded by other reasons? non-zero mass of the field? how big and dense would have to be a nuclear chunk of matter with non-zero chromodynamic dipolar moment, in order to produce a measurable macroscopic field?
 A: The nuclear strong force, which binds protons and neutrons to each other in an atomic nucleus, is essentially a higher-order multipole effect. So the effect does exist. However, it wouldn't extend to macroscopic lengths, because the carrier of such a force would be some combination of quarks and gluons which is necessarily massive, and therefore the force has a finite range.
For the strong nuclear force, you can use the one-pion exchange potential to do a simplified calculation.
$$V_\text{OPEP} = \hbar cC\ \vec\tau_1\cdot\vec\tau_2\biggl[\vec\sigma_1\cdot\vec\sigma_2 + S_{12}\biggl(1 + \frac{3}{mr} + \frac{3}{(mr)^2}\biggr)\biggr]\frac{e^{-rm_\pi c/\hbar}}{r}$$
where $C$ is a constant. I've found conflicting values for it: $C = \frac{g_\pi^2}{12}\bigl(\frac{m_\pi}{m_p}\bigr)^2 = 0.42$ with $g_\pi = 15$, or $C = \frac{m_\pi^2c^4}{12\pi}\bigl(\frac{g_A}{\sqrt{2}(132\text{ MeV})}\bigr)^2 = 0.023$, with $g_A = 1.25$, or perhaps something else. Anyway, the effective coupling $g_\pi$ or $g_A$ is basically related to the higher-order color multipoles of the proton or other system (I'm not sure if it's known exactly how), so in theory you could adjust those values to see how large you would need them to be to get a given effect.
