Lagerbaer has already answered OP's question for $n=4$ distinguishable spin dublets. More generally, the number of spin multiplets of $n$ distinguishable spin dublets 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 $\oplus$. Explicitly, the first few tensor powers read

$$ \underline{\large\bf 2}^{\otimes 1} ~=~\underline{\large\bf 2}, $$ $$ \underline{\large\bf 2}^{\otimes 2} ~=~\underline{\large\bf 3}~\oplus~\underline{\large\bf 1} ,$$ $$ \underline{\large\bf 2}^{\otimes 3} ~=~\underline{\large\bf 4}~\oplus~2~\underline{\large\bf 2} ,$$ $$ \underline{\large\bf 2}^{\otimes 4} ~=~\underline{\large\bf 5}~\oplus~3~\underline{\large\bf 3} ~\oplus~2~\underline{\large\bf 1} ,$$ $$ \underline{\large\bf 2}^{\otimes 5} ~=~\underline{\large\bf 6}~\oplus~4~\underline{\large\bf 4} ~\oplus~5~\underline{\large\bf 2} ,$$ $$ \underline{\large\bf 2}^{\otimes 6} ~=~\underline{\large\bf 7}~\oplus~5~\underline{\large\bf 5} ~\oplus~9~\underline{\large\bf 3}~\oplus~5~\underline{\large\bf 1} ,$$ $$ \vdots$$

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. The above pattern resembles Pascal's triangle. Clearly the general formula is of the form

$$ \underline{\large\bf 2}^{\otimes n}~=~\bigoplus_{k=0}^{[\frac{n}{2}]} m_{n,k} ~\underline{\large\bf n+1-2k}, $$

where the multiplicities $m_{n,k}\in \mathbb{N}_{0}$ satisfy  $$ m_{n,k}~=~0 \quad\text{for}\quad k> [\frac{n}{2}], $$

and

$$ m_{n,k-1}+ m_{n,k}~=~m_{n+1,k}. $$

A closed formula for the multiplicities reads (hattip:Trimok)

$$m_{n,k}~=~ \frac{n!~(n + 1 - 2k)}{k!~ (n + 1 - k)!}. $$

Lagerbaer has already answered OP's question for $n=4$ distinguishable spin dublets. More generally, the number of spin multiplets of $n$ distinguishable spin dublets 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 $\oplus$. Explicitly, the first few tensor powers read

$$ \underline{\large\bf 2}^{\otimes 1} ~=~\underline{\large\bf 2}, $$ $$ \underline{\large\bf 2}^{\otimes 2} ~=~\underline{\large\bf 3}~\oplus~\underline{\large\bf 1} ,$$ $$ \underline{\large\bf 2}^{\otimes 3} ~=~\underline{\large\bf 4}~\oplus~2~\underline{\large\bf 2} ,$$ $$ \underline{\large\bf 2}^{\otimes 4} ~=~\underline{\large\bf 5}~\oplus~3~\underline{\large\bf 3} ~\oplus~2~\underline{\large\bf 1} ,$$ $$ \underline{\large\bf 2}^{\otimes 5} ~=~\underline{\large\bf 6}~\oplus~4~\underline{\large\bf 4} ~\oplus~5~\underline{\large\bf 2} ,$$ $$ \underline{\large\bf 2}^{\otimes 6} ~=~\underline{\large\bf 7}~\oplus~5~\underline{\large\bf 5} ~\oplus~9~\underline{\large\bf 3}~\oplus~5~\underline{\large\bf 1} ,$$ $$ \vdots$$

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. The above pattern resembles Pascal's triangle. Clearly the general formula is of the form

$$ \underline{\large\bf 2}^{\otimes n}~=~\bigoplus_{k=0}^{[\frac{n}{2}]} m_{n,k} ~\underline{\large\bf n+1-2k}, \qquad n\in \mathbb{N}.$$

Here the multiplicities $m_{n,k}\in \mathbb{N}_{0}$ satisfy 

$$ m_{n,k}~=~0 \quad\text{for}\quad k> [\frac{n}{2}], $$

$$ m_{n,0}~=~1,$$

and

$$ m_{n,k-1}+ m_{n,k}~=~m_{n+1,k}\quad\text{for}\quad k \geq 1. $$

A closed formula for the multiplicities reads (hattip:Trimok)

$$m_{n,k}~=~ \frac{n!~(n + 1 - 2k)}{k!~ (n + 1 - k)!}. $$

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.

Lagerbaer has already answered OP's question for $n=4$ distinguishable spin dublets. More generally, the number of spin multiplets of $n$ distinguishable spin dublets 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 $\oplus$. Explicitly, the first few tensor powers read

$$ \underline{\large\bf 2}^{\otimes 1} ~=~\underline{\large\bf 2}, $$ $$ \underline{\large\bf 2}^{\otimes 2} ~=~\underline{\large\bf 3}~\oplus~\underline{\large\bf 1} ,$$ $$ \underline{\large\bf 2}^{\otimes 3} ~=~\underline{\large\bf 4}~\oplus~2~\underline{\large\bf 2} ,$$ $$ \underline{\large\bf 2}^{\otimes 4} ~=~\underline{\large\bf 5}~\oplus~3~\underline{\large\bf 3} ~\oplus~2~\underline{\large\bf 1} ,$$ $$ \underline{\large\bf 2}^{\otimes 5} ~=~\underline{\large\bf 6}~\oplus~4~\underline{\large\bf 4} ~\oplus~5~\underline{\large\bf 2} ,$$ $$ \underline{\large\bf 2}^{\otimes 6} ~=~\underline{\large\bf 7}~\oplus~5~\underline{\large\bf 5} ~\oplus~9~\underline{\large\bf 3}~\oplus~5~\underline{\large\bf 1} ,$$ $$ \vdots$$

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. The above pattern resembles Pascal's triangle. Clearly the general formula is of the form

$$ \underline{\large\bf 2}^{\otimes n}~=~\bigoplus_{k=0}^{[\frac{n}{2}]} m_{n,k} ~\underline{\large\bf n+1-2k}, $$

where the multiplicities $m_{n,k}\in \mathbb{N}_{0}$ satisfy $$ m_{n,k}~=~0 \quad\text{for}\quad k> [\frac{n}{2}], $$

and

$$ m_{n,k-1}+ m_{n,k}~=~m_{n+1,k}. $$

A closed formula for the multiplicities reads (hattip:Trimok)

$$m_{n,k}~=~ \frac{n!~(n + 1 - 2k)}{k!~ (n + 1 - k)!}. $$