To quantify the radio-frequency shielding properties of the aluminum sheet in this scenario, you can use analytical approximations and simulations. Let's break down the process of the analytical approximations.
First, you can use some analytical approximations to get a rough idea of the shielding effectiveness of the aluminum sheet. The shielding effectiveness \$\text{SE}\$ is usually measured in decibels \$\left[\text{dB}\right]\$ and represents the reduction in electromagnetic field strength when passing through a material. It's given by:
$$\left|\text{SE}\right|=\left|20\log_{10}\left(\frac{\displaystyle\text{E}_\text{i}}{\displaystyle\text{E}_\text{t}}\right)\right|\tag1$$
Where \$\displaystyle\text{E}_\text{i}\$ is the incident field strength (without the shield) and \$\displaystyle\text{E}_\text{t}\$ is the transmitted field strength (with the shield).
For an infinite sheet of metal, the shielding effectiveness can be estimated using the "Skin Depth" \$\delta\$ at the frequency of interest. The skin depth represents how deeply electromagnetic waves can penetrate a conductor and is given by:
$$\delta=\frac{\displaystyle1}{\displaystyle\sqrt{\pi\mu_0}}\cdot\frac{\displaystyle1}{\displaystyle\sqrt{\sigma\text{f}}}=\frac{\displaystyle500\sqrt{10}}{\pi}\cdot\frac{\displaystyle1}{\displaystyle\sqrt{\sigma\text{f}}}\tag2$$
Where \$\mu_0=4\pi\cdot10^{-7}\space\text{H/m}\$ is the vacuum magnetic permeability, \$\sigma\$ is the conductivity of the material (for aluminum, approximately \$3.5\cdot10^7\space\text{S/m}\$) and \$\text{f}\$ is the frequency in \$\left[\text{Hz}\right]\$.
Once you have the skin depth, you can use it to approximate the fraction of the incident field that gets through the sheet:
$$\frac{\displaystyle\text{E}_\text{i}}{\displaystyle\text{E}_\text{t}}=\exp\left(-\frac{\displaystyle\text{d}}{\displaystyle\delta}\right)\tag3$$
Where \$\text{d}\$ is the thickness of the sheet.
Now you can plug this fraction into the shielding effectiveness formula to get an approximate shielding effectiveness in dB. So, we end up with:
$$\left|\text{SE}\right|=\frac{\displaystyle20}{\displaystyle\ln\left(10\right)}\cdot\frac{\displaystyle\pi\text{d}\sqrt{\sigma\text{f}}}{\displaystyle500\sqrt{10}}\approx0.01725814\text{d}\sqrt{\sigma\text{f}}\tag4$$
