Trend in the boiling point of the hydrogen halides
There are formulae you can use to calculate the strength of interactions between two molecules, which are derived from first principles. You can find an overview here on Chemistry LibreTexts. The two relevant ones are:

$$U_{\mathrm{dipole}} = -\frac{2}{3kT} \frac{p^4}{(4\pi\varepsilon_0)^2r^6}$$

$$U_{\mathrm{dispersion}} = -\frac{3}{4} \frac{\alpha^2 I}{(4\pi\varepsilon_0)^2r^6}$$


$p$ is the molecular dipole moment
$r$ is the average separation between neighbouring molecules
$\alpha$ is the polarisability of the molecule, which in a nutshell, measures how easy it is to create an induced dipole in the molecule
$I$ is the first ionisation energy of the molecule


If you plug all the numbers in for the hydrogen halides $\ce{HX}$, you will find that the magnitude of the dispersion forces is much larger than the magnitude of the dipole-dipole attractions, i.e. $\left|U_{\mathrm{dispersion}}\right| \gg \left|U_{\mathrm{dipole}}\right|$.

Unfortunately, I don't have all the necessary data on hand, but I can quote some results that were given in first-year lecture notes at Oxford (which should be legitimate). For $\ce{HCl}$, dispersion forces contribute $86\%$ to the intermolecular attractions, and for $\ce{HI}$, they contribute $99\%$.

Considering this fact, it is not surprising that variations in the magnitude of dispersion forces affect the boiling point much more than variations in the magnitude of the dipole-dipole attractions. If you had the all the data required and evaluated the interactions using the two formulae above, you would reach the same conclusion quantitatively.

In fact, this trend is hardly unique to the hydrogen halides. The hydrides of the Group 15 and Group 16 elements behave exactly the same way:

$$\begin{array}{|cc|cc|}
\hline
\text{Species} &amp; \text{Boiling point / }\mathrm{^\circ C} &amp; \text{Species} &amp; \text{Boiling point / }\mathrm{^\circ C} \\
\hline
\ce{PH3} &amp; -87.5 &amp; \ce{H2S} &amp; -60.3 \\
\ce{AsH3} &amp; -62.4 &amp; \ce{H2Se} &amp; -41.3 \\
\ce{SbH3} &amp; -18.4 &amp; \ce{H2Te} &amp; -4 \\
\hline
\end{array}$$
(source: Greenwood &amp; Earnshaw, Chemistry of the Elements 2nd ed., pp 557, 767)

Of course, the first-row hydrides are left out of the discussion because of hydrogen bonding, which makes their boiling points anomalously high.