Every system with $N$ or more electrons lies in a Hilbert space $H=H_{\text{space}} \otimes H_{\text{spin}}$, with $H_{\text{space}}=H_{\text{space}}^{1}\otimes\cdots\otimes H_{\text{space}}^{N}$ and $H_{\text{spin}}=H_{\text{spin}}^{1}\otimes\cdots\otimes H_{\text{spin}}^{N}$, $H^i$ being the $i$-th particle space. So the system has a state $|\Psi\rangle = |\Psi\rangle_{\text{space}} \otimes |\Psi\rangle_{\text{spin}} \in H$.
What I couldn't come up with is an antisymmetric spin ket $|\Psi\rangle_{\text{spin}}$ when there was more than 3 electrons. This would mean that the only way of antisymmetrizing $|\Psi\rangle$, for $N\geq 3$, is by antisymmetrizing only the spatial part. I think it's strange, since for $N=2$ we do have an antisymmetric spin ket (the singlet state), so why wouldn't be such kets for $N\geq 3$?
Ignoring the spatial part, and assuming $N\geq 3$, if we want to describe $N$ identical spins $\sigma_k = \pm$, we need to antisymmetrize the ket $|\sigma_1\rangle |\sigma_2\rangle \cdots |\sigma_N\rangle$ in the following way
\begin{equation} |\Psi\rangle_{\text{spin}} = \frac{1}{\sqrt{N}}\sum_{p \in S_N}sg(p)|\sigma_{p(1)}\rangle |\sigma_{p(2)}\rangle \cdots |\sigma_{p(N)}\rangle \end{equation}
Let's take, for example, the following ket (which we want to antisymmetrize)
\begin{equation} |\phi\rangle=\underbrace{|+\rangle\cdots|+\rangle}_{n}\underbrace{|-\rangle\cdots|-\rangle}_{m} \quad (n+m=N) \end{equation}
If we only look at the permutations $p$ which don't change $|\phi\rangle$, we end up with a subgroup $S_n S_m\subset S_N$, made up of:
$S_n$ = permutations $\alpha\in S_N$ which don't change the "$\underbrace{|+\rangle\cdots|+\rangle}_{n}$" part and don't touch the "$\underbrace{|-\rangle\cdots|-\rangle}_{m}$" part
and:
$S_m$ = permutations $\beta\in S_N$ which don't touch the "$\underbrace{|+\rangle\cdots|+\rangle}_{n}$" part and don't change the "$\underbrace{|-\rangle\cdots|-\rangle}_{m}$" part
With $S_n S_m$ being all the permutations of the form $\alpha \circ \beta$
But the thing is that half of the elements of $S_n$ are even and the other half are odd, so the following sum is zero:
\begin{align} A(|\phi\rangle) \stackrel{\text{def}}{=} &\sum_{p\in S_n S_m} sg(p) |\phi\rangle = \sum_{\alpha\beta\in S_n S_m} sg(\alpha\beta) |\phi\rangle = \sum_{\alpha\beta\in S_n S_m} sg(\alpha)sg(\beta) |\phi\rangle = \\ = &\sum_{\alpha\in S_n}\sum_{\beta\in S_m} sg(\alpha)sg(\beta) |\phi\rangle = \underbrace{\left(\sum_{\alpha\in S_n} sg(\alpha)\right)}_{0} \sum_{\beta\in S_m}sg(\beta) |\phi\rangle = 0 \end{align}
And a similar computation could have been done to every permutation of $|\phi\rangle$, so, noticing that the original ket $|\Psi\rangle_{\text{spin}}$ is a sum of terms like $A(|\phi'\rangle)$, with $|\phi'\rangle$ being permutations of $|\phi\rangle$ which do change it (unlike before), it turns out that $|\Psi\rangle_{\text{spin}} = 0$ for every $N>2$ ! (with $|\Psi\rangle_{\text{spin}}$ antisymmetric)
