What is meant by complete outer shell? Why do the noble gases have zero valency?
When we refer to "outer shells" we are talking about the highest n-level, but also limiting ourselves to the s- and p-orbitals. The reason for this has to do with how the effective nuclear charge "felt" by the valence electrons changes as you move through successively higher energy configurations.

In summary, $s$- and $p$- electrons are screened less effectively by inner shells, and so for a given n-value, the $s$ and $p$ orbitals of the next n fill before the current n (they have lower energy).

In your example using argon, this means that the 3d shell actually has no electrons - the condensed configuration is:

$$
\ce{Ar: \space [Ne]}3s^23p^6
$$

For the next noble gas, krypton, the condensed electron configuration is:

$$
\ce{Kr: \space [Ar]}4s^23d^{10}4p^6
$$

Note that it has $3d$ electrons at a higher energy than what argon had, but since the value of n roughly corresponds to the distance of the orbital from the nucleus, the $3d$ orbitals are physically closer to the nucleus than the $4s$ and $4p$ orbitals. In other words, for single atoms, the lowest energy electron configuration always fills the s- and p- orbitals first, and so when we are looking  for the valence or "outermost" electrons, that is where we need to look - we sort of ignore the $d$- and $f$-orbital electrons for these kinds of problems.