Kernel of skew-symmetric matrix of rank $n-1$ with $n$ odd: is this a known result?

$\def\pf{\mathop{\rm pf}}$I do not know the reference, and it seems that it is easier to prove the claim than to find a reference ;).

The proof follows the lines of the proof of a similar formula for determinants. Let $M=[m_{ij}]_{i,j=1}^n$. Denote by $M_i$ the matrix obtained from $M$ by deleting the $i$th row and the $i$th column. Let $\overline M_i$ be the $(n+1)\times(n+1)$ skew-symmetric matrix obtained from $M$ by doubling the $i$th row and the $i$th column and putting the copies to the top and to the left, respectively. We have $\det \overline M_i=0$ and hence $\pf \overline M_i=0$.

On the other hand, in view of the formula $$ \pf A=\sum_{j=2}^{2k}(-1)^j a_{1j}\pf A_{\hat 1\hat j} $$ (for a skew-symmetric $2k\times 2k$ matrix $A=[a_{ij}]$), we have $$ 0=\pf \overline M_i=\sum_{j=1}^nm_{ij}\cdot (-1)^{j+1}\pf M_j, $$ which shows that the $i$th element in the product of $M$ by your column is zero.

Another proof:

Let $\omega\in\Lambda^2(V^*)$ correspond to the skew-symmetric matrix $M$, i.e. $\omega = \sum_{i<j} M_{ij} f_i\wedge f_j$ where $f_i$ is the canonical basis of $(\mathbb{R}^n)^*$. Let $\eta = f_1\wedge\cdots\wedge f_n$ be the canonical volume form on $\mathbb{R}^n$. Then $v$ is uniquely defined by $\iota_{v}\eta = \omega\wedge\cdots\wedge\omega$ ($\frac{n-1}{2}$ terms). Now that we have defined everything in a coordinate-free way, we can choose a basis where

$M=\left(\begin{array}{cccccc}0&\lambda_1&&&&\\-\lambda_1&0&&&&\\&&\ddots&&&\\&&&0&\lambda_\frac{n-1}{2}&\\ &&&-\lambda_\frac{n-1}{2}&0&\\&&&&& 0\end{array}\right)$

The change of basis can be taken to be in $SO(n)$, in particular the expression for $\eta$ does not change.

In this case, deleting anything but the last row replaces one of the blocks by a $0$, and the resulting Pfaffian vanishes by inspection. Thus only the last component of $v$ does not vanish, so $v$ is in the kernel of $M$.