What is the energy density of the zero Higgs field in SI units? Higgs boson has a mass, and a vacuum expectation value, both are in electronvolts.
An uncommon feature of the Higgs field is that to its zero energy density belongs a non-zero field value.
Is it possible, probably with QFT, to calculate its energy density (per unit volume)? In layman terms, how would it work?
 A: The "Mexican hat" Higgs potential is
$$V(H)=\lambda(|H|^2-v^2)^2$$
where $v$ is the Higgs vacuum expectation value, 246 GeV, and the dimensionless Higgs coupling $\lambda$ is about 0.0323.
To get this value for $\lambda$, look at the quadratic "mass term" when expanding the Higgs field around its VEV, $H=v+h$. It is $4\lambda v^2 h^2$, so the mass of the Higgs is $m_H^2=8\lambda v^2$ and thus $\lambda=m_H^2/8v^2$. Using 125 GeV as the measured Higgs mass, one gets $\lambda\approx 125^2/(8\times 246^2)$.
When the Higgs field $H$ is zero, the potential is $\lambda v^4$, which is about $1.18\times 10^8\text{ GeV}^4$. The conversion for energy density between natural units and SI units is
$$1\text{ GeV}^4\approx 2.08\times10^{37}\text{ J/m}^3$$
so the zero-Higgs-field energy density is about $2.45\times 10^{45}\text{ J/m}^3$. The corresponds to a mass density of $2.72\times 10^{28}\text{ kg/m}^3$. This is about 10 orders of magnitude higher than density at the center of a neutron star, but about 68 orders of magnitude lower than the Planck density.
Note that this energy density is not the Higgs energy density now, when $H=v$. That energy density is, as you said, zero.
