Size bound on Büchi automaton for complement
Once you have the bound $2^{2n^2}$ on the number of classes, you can note that each state in the complement automaton corresponds to an $\omega$-regular language of the form $L_{f}L_{g}^\omega$, with $L_f$ and $L_g$ being functions from $Q$ to $2^Q\times 2^Q$. 

Each such language is determined by $L_f$ and $L_g$, and there are $2^{2n^2}$ options for these functions, so you get a total of $(2^{2n^2})^2=2^{4n^2}$ options.

UPDATE FOLLOWING COMMENT:
The correspondence of the number $2^{4n^2}$ to the states of the constructed automaton is not exact. Observe that for every language $L_f$, we can construct a nondeterministic NFA $A_f$ (as is done in Theorem 1 in the Wikipedia link you posted). Moreover, the size of $A_f$ is polynomial in the size of the original NBW $A$. Thus, the size of an NBW for $L_fL_g^\omega$ is polynomial. You then take a union of $2^{4n^2}$ such automata, and obtain the complement NBW, so the size of the complement NBW is:
$$2^{4n^2}\cdot poly(n)=2^{4n^2 +\log poly(n)}=2^{O(4n^2)}$$
So it's a little less tight than $O(2^{4n^2})$, but it's still only singly exponential (perhaps there is a tighter analysis, but I'm not aware of one).