How to calculate the argument of perihelion for exoplanets with incomplete orbital elements?
Ford et al. (2008) give the following relationship for eccentric transiting planets:

$$\frac{t_\mathrm{D}}{P} \simeq \frac{R_\ast}{\pi a \sqrt{1-e^2}}\sqrt{(1+r)^2 - b^2} \left(\frac{d_\mathrm{t}}{a}\right)$$

Where $t_\mathrm{D}$ is the total transit duration, $P$ is the orbital period, $R_\ast$ is the stellar radius, $a$ is the semimajor axis, $e$ is the eccentricity, $r \equiv R_\mathrm{p}/R_\ast$ is the ratio of planetary radius $R_\mathrm{p}$ to the stellar radius, $b \equiv d_\mathrm{t} \cos i / R_\ast$ is the impact parameter, $i$ is the inclination and $d_\mathrm{t} = a(1-e^2)/(1+e \cos \omega)$ is the distance between the star and the planet during the transit. By inverting this you would be able to estimate $\cos \omega$ from the parameters listed on the site, which unfortunately won't give you a unique value (and if you're unlucky, the various uncertainties may conspire to give you a value $\left|\cos \omega \right|&gt;1$).

In the case of Kepler-444, the website gives a link to the paper where the orbits were published, Campante et al. (2015). Their table 4 tabulates both $e \cos \omega$ and $e \sin \omega$ for the planets. Unfortunately it is not possible to simply determine the value from these estimates without having access to the full distribution of samples, so this isn't entirely helpful.

The good news is that the statistical significance of the non-zero eccentricities doesn't appear to be too high in this case, so you can probably treat the orbits as circular and not worry too much about it: tightly-packed systems like this usually have very low eccentricities otherwise they would become unstable.