It turns out that there's a pretty detailed analysis of this in Jackson, Classical Electrodynamics, sections 3.13, 5.13, and 9.5. Although Jackson does it in excruciating detail using Bessel functions and infinite series, it's actually pretty easy to pull out the basic ideas.
Start by considering a simpler problem. A thin, conducting sheet in the $x-y$ plane has a circular hole in it of radius $a$. Suppose that at large distances above the sheet, the electric field is uniform with magnitude $E_0$ and is in the $z$ direction, but the field is zero below the sheet. Then the field closer to the hole can be broken down into two terms: one for the field you'd have if the hole wasn't there, and another term that exists because the hole is there. At sufficiently large distances, the second term can be approximated as that of an electric dipole $p$, which by symmetry must be in the $z$ direction, and by linearity must be proportional to $E_0$. On dimensional grounds, we must have $p \propto E_0 a^3$. (The unitless constant of proportionality turns out to be $1/3\pi$, but that doesn't really matter for my purposes.)
A similar analysis holds for a magnetic field in the $x$ direction, with a magnetic dipole appearing at the hole. (The constant of proportionality is $2/3\pi$.)
Now consider an electromagnetic wave with wavelength $\lambda$ coming down from above. If $\lambda$ is large compared to $a$, then the electrostatic and magnetostatic results above still hold at any given time. Therefore we have dipole radiation coming from the hole, with amplitude proportional to $a^3$ and power $P$ proportional to $a^6$. The power $P_0$ incident on the hole is proportional to $a^2$, so the fraction of the power transmitted $T=P/P_0$ is proportional to $a^4$. On dimensional grounds, we must therefore have $T\propto (a/\lambda)^4$, where the constant of proportionality is unitless. This also makes sense because the power radiated by a dipole is proportional to $\omega^4$.
I think it's pretty straightforward to see how this would play out for surface waves on water. I would expect a narrow hole of width $h$ to act as a monopole source, which would radiate power proportional to $\omega^2$. Therefore the transmission must go like $\lambda^{-2}$, and on dimensional grounds we must therefore have $T\propto (h/\lambda)^2$.
All of this follows directly from two very simple considerations: (1) dimensional analysis, and (2) the proportionality of the amplitude of $\ell$-pole radiation to $\omega^{2(\ell+1)}$. Huygens' principle is never needed, and in fact would not have been valid in an even number of spatial dimensions (which is what we have in the case of the water waves).
This is all for a thin sheet. As noted by CuriousOne in a comment, when the sheet is thicker, you can treat the hole like a waveguide. I tried to extract the basic ideas from this web page: http://www.cvel.clemson.edu/emc/tutorials/Shielding02/Practical_Shielding.html The basic idea seems to be that the wave falls off exponentially, with a characteristic length $ L \propto a/ \sqrt{1-((2a)/ \lambda)^2}$ I've approximated $\lambda_c \approx 2a$ as the cutoff frequency, and the dependence on the details of the geometry (circular cross-section, rectangular,...) is all built into a unitless constant of proportionality, which I think is just the quantity $T$ previously discussed in this answer. So in the limit of long wavelength, you basically get exponential attenuation through the thickness of the wall, with characteristic length of order $a$.
In the long-wavelength limit, where the field acts like DC, it might also be possible to compare all of this to a Faraday ice pail.
