TAI is a weighted average of the clocks. The lower the estimated uncertainty of a clock, the more it contributes to the average. This is given in the second to last column in the first table of Circular T [1]. For example, in the last reporting period, Germany's PTB reported the lowest uncertainty, and therefore contributed most. Probably both their cesium fountains were running well. So a single much better clock would immediately improve the quality of TAI.

[1] https://webtai.bipm.org/ftp/pub/tai/Circular-T/cirt/cirt.409

but how do you estimate the uncertainty without another clock to compare against? It seems like a circular problem.... how do you prove you built a better clock without already having a better clock?

Estimating the uncertainty is indeed quite difficult.

Stability can be measured by comparing two uncorrelated clocks. This is why standard institutes tend to build them in pairs.

Measuring absolute accuracy is a different beast, since two similar clocks will also have similar systematic errors. Of course, one can estimate the magnitude of these errors, but there can always be unknown contributions. It is therefore useful to compare different implementations, for example cesium fountain clocks built at different institutes. Still, there can always be some conceptual problem that affects all clocks in a similar way.

As a general rule, sampling a normal distribution N times decreases the uncertainty by a multiple of 1/sqrt(N). For instance, to know the mean of an unknown distribution to within N digits, the number of observations must be on the order of 2N digits long. I learned this the hard way when attempting some statistical experiments; it may answer your question here.

How does that work for the median (as opposed to the mean)? Asking because median rejects outliers better than the mean.

In large samples the error bars for the median are sqrt(pi/2) times the error bars for the mean if the underlying distribution is normal (the link gives more general results, but that helps give some intuitions about being directly comparable)

https://en.wikipedia.org/wiki/Median#Sampling_distribution

It works the same way. Nine times the clocks, three times smaller standard deviation for both the mean and the median of their readings. In a normal (and any symmetric) distribution these two are the same anyway, so their difference will tend to zero as you increase the number of clocks and the measurements from a single reading of all the clocks look more and more like a bell curve. This is asymptotic behavior so the "rejects the outliers better" aspect does not come into play, and I do not know whether it is useful in this situation or not.