Let $l$ be even and $v \in \bigwedge^\ell \mathbb{C}^{\ell n}$. What conditions on vector $v$ imply $v^{\wedge n} \neq 0$?
Note, that for $\ell = 2$ the criteria is that $v$ must be non-degenerate (skew-symmetric bilinear form). Is there any reference of idea how one can approach at least the implication "condition on $v$ $\implies$ $v^{\wedge n}$ not nulls"?
Let $V$ be a complex vector space of dimension $\ell n$, where $\ell$ is an even positive integer. I will work with $\bigwedge^{\ell} V^*$ rather than $\bigwedge^{\ell} V$ so that we may think of alternating multilinear functions rather than the exterior algebra itself.
Recall that if we let $$ \text{Sh}(a,b) = \{\sigma\in S_{a+b}: \sigma(i)<\sigma(j) \text{ if } i< j\leq a \text{ or } a+1\leq i < j\} $$ so that if $\beta \in \bigwedge^{a} V^*, \gamma \in \bigwedge^b V^*$ then $$ \tag{$\dagger$} (\beta\wedge \gamma)(v_1,\ldots,v_{a+b}) = \sum_{\sigma \in \text{Sh}(a,b)}\text{sgn}(\sigma)\beta(v_{\sigma(1)},\ldots,v_{\sigma(a)}\gamma(v_{\sigma(a+1)},\ldots,v_{\sigma(a+b)}) $$
Updated Claim: If $\alpha \in \bigwedge^{\ell} V^*$ then $\alpha^n\neq 0$ then there exists a direct sum decomposition $V = \bigoplus_{k=1}^n W_k$ with $\dim(W_k)=\ell$ for all $k$, $(1\leq k \leq n)$ such that if $\iota_k\colon V\to W_k$ denotes the inclusion of $W_k$ into $V$ then $\iota_k^*(\alpha) = \alpha_{|W_k} \in \bigwedge^\ell W_k$ is nonzero for each $k$, $1\leq k\leq n$.
Proof: If $\alpha\in \bigwedge^\ell V^*$ and $\alpha^n\neq 0$, then certainly $\alpha^{n-1} \in \bigwedge^{(n-1)\ell}V^*\neq 0$, hence there must be a set of vectors $B_1 = \{v_1,\ldots,v_{(n-1)\ell}\} \subseteq V$ with $\alpha^{n-1}(v_1,\ldots,v_{(n-1)\ell})\neq 0$. Since $\alpha^{n-1}$ is alternating, it follows that $B_1$ is a linearly independent set, hence we can extend it to a basis of $V$, that is, we can find $B_2 = \{v_{\ell(n-1)+1},\ldots, v_{\ell n}\}$ such that $B_1\sqcup B_2$ is a basis of $V$.
Thus using ($\dagger$) we see that $$ \begin{split} 0 &\neq \alpha^n(v_1\ldots,v_{\ell n}) \\ &= \alpha^{n-1}\wedge \alpha(v_1,\ldots v_{\ell n}) \\ &= \sum_{\sigma\in \text{Sh}(\ell(n-1),\ell)}\text{sgn}(\sigma)\alpha^{n-1}(v_{\sigma(1)},\ldots,v_{\sigma(\ell(n-1))})\alpha(v_{\sigma(\ell(n-1)+1)},\ldots,v_{\sigma(\ell n)}) \end{split} $$
and thus it follows that there must exist at least one $\sigma \in \text{Sh}(\ell(n-1),\ell)$ such that if we set $u_i = v_{\sigma(i)}$ then $\alpha^{n-1}(u_1,\ldots,u_{\ell(n-1)})$ and $\alpha(u_{\ell(n-1)+1},\ldots,u_{\ell n})$ are both nonzero. Thus if $V_1 = \text{span}\{u_1,\ldots,u_{\ell(n-1)}\}$ and $W_1 = \text{span}\{u_{\ell(n-1)+1},\ldots,u_{\ell n}\}$, and $\iota_1\colon W_1 \to V$, $j_1\colon V_1 \to V$ denote the inclusion maps, $\iota_1^*(\alpha) \neq 0$, and $j_1^*(\alpha^{n-1})\neq 0$. But since $j_1^*(\alpha^{n-1})= (j_1^*(\alpha))^{n-1}$, we may apply induction to the pair $(V_1, j_1^*(\alpha))$ to complete the proof of the necessity of the condition.
The converse is false:
Suppose that $V= \mathbb C^8$ and $\{f_1,\ldots,f_8\}$ is the dual basis of the standard basis of $\mathbb C^8$. Write $f_{ij...}$ for $f_i\wedge f_j \wedge \ldots$. Then if $\ell=4$ and $n=2$, if we let $\alpha = f_{1234} + f_{1278} + f_{3456} - f_{5678}$, then taking $W_1 = \text{span}\{e_1,e_2,e_3,e_4\}$ and $W_2 = \{e_5,e_6,e_7,e_8\}$ we see that $\alpha_{|W_1}\neq 0$ and $\alpha_{|W_2} \neq 0$, but $$ \begin{split} \alpha^2 &= (f_{1234} + f_{3456}+f_{1278} - f_{5678})^2 \\ &= -2f_{1234}\wedge f_{5678} + 2f_{3456}f_{1278} \\ &= -2f_{12345678} + 2f_{12345678} = 0. \end{split} $$ Of course the issue is that the subspaces $W_1' = \text{span}\{e_3,e_4,e_5,e_6\}$ and $W_2' = \{e_1,e_2,e_7,e_8\}$ also satisfy the conclusion of the claim, and the two top forms the two pairs contribute cancel each other.
Of course if $V= \bigoplus_{k=1}^n W_k$ and $p_k\colon V \to W_k$ the associated projection maps, then if $\alpha= \sum_{k=1}^n p_k^*(\omega_k)$ where $\omega_k \in \bigwedge^{\ell}(W_k)\backslash \{0\}$ it follows that $\alpha^n\neq 0$, but I don't at the moment see an obvious reason why this sufficient condition should be necessary.