Lagerbaer has already answered OP's question for $n=4$. More generally, the number of spin multiplets of $n$ distinguishable spin half states can be deduced from repeated applications of the $SU(2)$ Clebsch-Gordan fusion rule
$$ \underline{\large\bf 2} \otimes \underline{\large\bf n}~=~ \left\{ \begin{array}{lcl} \underline{\large\bf n+1}\oplus\underline{\large\bf n-1} &\text{for}& n\geq 2, \\ \underline{\large\bf n+1}&\text{for}& n=1, \end{array} \right. $$
and the distributive law for $\otimes$ and $\oplus$. The pattern resembles Pascal's triangle. Explicitly, the first few terms in the tensor product read
$$ \underline{\large\bf 2}^{\otimes n}~=~\underline{\large\bf n+1} ~\oplus~\frac{n-1}{1!}\underline{\large\bf n-1} ~\oplus~\frac{n(n-3)}{2!}\underline{\large\bf n-3} $$ $$~\oplus~\frac{n(n-1)(n-5)}{3!}\underline{\large\bf n-5} ~\oplus~\frac{n(n-1)(n-2)(n-7)}{4!}\underline{\large\bf n-7} ~\oplus~\ldots $$
Here the irreps $\underline{\large\bf 1}$, $\underline{\large\bf 2}$,$\underline{\large\bf 3}$, $\ldots$, denote singlet, dublet, triplet, $\ldots$, i.e., spin $0$, $\frac{1}{2}$, $1$, $\ldots$, respectively.