If you identify a vector $v \in \mathbb{R}^n$ with the linear mapping $$ F_v \,:\, \mathbb{R}^n \to \mathbb{R}^n \,:\, x \to (v^T\cdot x, 0, \ldots 0)^T \text{,} $$ i.e. if you identify the vector $v=(v_1,\ldots,v_n)$ with the matrix $$ F_v = \begin{pmatrix}v_1 &\ldots &v_n\\ 0&\ldots&0\\\vdots&\ddots&\vdots\\0&\ldots&0\end{pmatrix} \text{,} $$ you can say that $$ \det v = \det F_v = \begin{cases} v_1 &\text{if $n=1$} \\ 0 &\text{otherwise.} \end{cases} $$ Though whether that's particularly helpful is a different question..

Determinants are defined for square matrices, only. If you are trying to use the determinant of a vector (and $n\neq1$) then you are somehow on a very wrong track.

Well, the determinant is defined not as a function of a vector, but rather as a multilinear function of $n$ vectors in $n$-space, in other words $\det : \prod_{i=1}^n \mathbb{R}^n\to\mathbb{R}$ and is multilinear. In this setup it doesn't make much sense talking about $\det(v)$ for some $v\in \mathbb{R}^n$, because $\det$ was built to receive as it's input $n$ vectors instead of just one.

If you insist in defining the determinant of a vector however, one possibility would be to talk about the new function obtained defining $f_v: \prod_{i=1}^{n-1} \mathbb{R}^n \to \mathbb{R}$ by setting:

$$f_v(w_1, \dots, w_{n-1})=\det(v, w_1\dots, w_{n-1})$$

So that we would me mapping $v \mapsto \det(v, \cdot, \dots, \cdot)$, that's probably a common approach when we deal with tensors in general: if we understand tensors as multilinear maps, then if we give just one argument to the tensor it gives us back a multilinear map that can receive one less input, because one of the inputs was fixed, in that framework $\det$ is a tensor and giving just one vector to it gives back a new function of $n-1$ vectors.