WEll I have very little idea but this is where my play took me.
How many subsets of S are there altogether. This is the number of ways that A can be chosen
\(\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\)
subsets b and C can be shosen the same way so
number of ways subsets A,B and C can be chosen is \(\left[\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\right]^3\)
But what if A is a subset of B and B is a subset of C ...
Ways to chose C is \(\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\)
then
ways to chose B is \(\displaystyle \sum_{m=0}^k\;\;\binom{k}{m} \qquad m\le k\)
ways to choose A is \(\displaystyle \sum_{t=0}^m\;\;\binom{m}{t}\qquad t\le m\)
Put thse toghether and I get the number of ways \(A \subseteq B \subseteq C\)
is \( \displaystyle \left[ \sum_{k=0}^n\;\;\; \binom{n}{k}\left[\sum_{m=0}^k\;\;\binom{k}{m}\left[\sum_{t=0}^m\; \binom{m}{t}\right]\right]\right]\\~\\ \) that is probably not displayed in the right order. 
So prob would be
\( \frac{\displaystyle \left[ \sum_{k=0}^n\;\;\; \binom{n}{k}\left[\sum_{m=0}^k\;\;\binom{k}{m}\left[\sum_{t=0}^m\; \binom{m}{t}\right]\right]\right]\\~\\ }{\left[\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\right]^3} \)
I don't know how to display that properly, the order and the brackets are most likely wrong.......
