4444

Let $T(n)$ mean the number of $n$ 's digits. For example: $T(2017)=4$ .

Since $T(4444^{4444})<T(10000^{4444})=T((10^4)^{4444}=10^{17776})=17777$ , it is clear that $S(4444^{4444})<17777*9=159993$ .

Therefore $S(S(4444^{4444}))<S(99999)$ , because the sum of the digits of $99999$ is at least as big as the $S(.)$ of any other number less than $159993$ . $S(99999)=45$ .

From that $S(S(S(4444^{4444})))$ is at most $S(39)=12$ . (Because $39$ has the highest $S(.)$ of any number less that $45.$ )

We also know that if we divide $S(S(S(4444^{4444})))$ by $9$ , the remainder is the same as the remainder upon dividing $4444^{4444}$ by $9$ .

Since $4444\equiv7 \ \mod9$ , and $7^3=343\equiv1 \ \mod9$ , we get $4444^{4444}\equiv7^{4444}=7*7^{4443}=7*(7^3)^{1481}\equiv7 \ \mod9$ .

The answer must be $\boxed{7}$ .