Using the determinant is something of a distraction. In completely coordinate-free terms, and after applying the observation in abx's answer, the map you're trying to understand is the wedge product map
$$V^k \ni (v_1, ..., v_k) \mapsto v_1 \wedge ... \wedge v_k \in \Lambda^k(V)$$
where $V$ is a finite-dimensional vector space (of dimension, say, $n$). So the variety you're trying to understand is the variety of pure tensors sitting inside $\Lambda^k(V)$.
This variety has definitely been studied classically. The pure tensors are characterized by a set of quadratic relations called the Plücker relations. They cut out a subvariety of the projective space $\mathbb{P}(\Lambda^k(V))$ into which the Grassmannian $\text{Gr}_k(V)$ embeds via the Plücker embedding
$$\text{Gr}_k(V) \ni \text{span}(v_1, ..., v_k) \mapsto v_1 \wedge ... \wedge v_k \in \mathbb{P}(\Lambda^k(V)).$$
It follows that the pure tensors are the affine cone of $\text{Gr}_k(V)$ in this embedding, and so the space of pure tensors has dimension one greater than $\text{Gr}_k(V)$, hence $k(n-k) + 1$.
Example. The first nontrivial Plücker relation occurs when $\dim V = 4$ and $k = 2$. We can describe a pure tensor $v \wedge w$ using a $2 \times 4$ matrix, say
$$\left[ \begin{array}{ccc} c_1 & c_2 & c_3 & c_4 \\ d_1 & d_2 & d_3 & d_4 \end{array} \right],$$
where the first row gives the components of $v$ in some basis and the second row gives the components of $w$. The components of $v \wedge w$ are then given by the $2 \times 2$ minors of this matrix, and the only nontrivial Plücker relation, given by expanding $(v \wedge w) \wedge (v \wedge w) = 0$, is
$$p_{12} p_{34} - p_{13} p_{24} + p_{14} p_{23} = 0$$
where $p_{ij} = \det \left[ \begin{array}{cc} c_i & c_j \\ d_i & d_j \end{array} \right]$. Hence the pure tensors have dimension $6 - 1 = 2(4 - 2) + 1 = 5$ in this case.
Identifying forms and vector fields is a somewhat different matter. The canonical identifications are as follows. As abx says, the wedge product gives a nondegenerate pairing $\Lambda^k(V) \otimes \Lambda^{n-k}(V) \to \Lambda^n(V)$. The partial determinant gives a nondegenerate pairing $\Lambda^k(V) \otimes \Lambda^k(V^{\ast}) \to 1$ ($1$ the ground field) which allows us to commute taking duals with taking exterior powers. This gives identifications
$$\Lambda^k(V) \cong \Lambda^n(V) \otimes \Lambda^{n-k}(V)^{\ast} \cong \Lambda^n(V) \otimes \Lambda^{n-k}(V^{\ast}).$$
Hence to identify $\Lambda^k(V)$ with $\Lambda^{n-k}(V^{\ast})$ requires precisely the data of a nonzero element of $\Lambda^n(V)$. Globally this says that to identify vector fields with $n-1$-forms on an $n$-dimensional manifold requires precisely the data of a volume form. In particular you do not need a metric.
