The $n = 3$ case is a straightforward computation using the identification of the cross product on $\mathbb{R}^{3}$ with the Lie bracket on $\mathfrak{so}(3)$. Namely, defined $\hat{x}:\mathbb{R}^{3} \to \mathbb{R}^{3}$ by $\hat{x}y = x \times y$. Then $|\hat{x}|^{2} = 2|x|^{2}$, and the Jacobi identity implies $[\hat{x}, \hat{y}] = \widehat{x \times y}$, so $2|[\hat{x}, \hat{y}]|^{2} = 2|\widehat{x \times y}|^{2} = 4|x \times y|^{2} \leq 2|x|^{2}|y|^{2} = |\hat{x}|^{2}|\hat{y}|^{2}$. 

For $n > 3$, the optimal result is not as straightforward. 
(For $n = 4$ it can be deduced from the $n = 3$ case using the special isomorphism $\mathfrak{so}(4) \simeq \mathfrak{so}(3) \times \mathfrak{so}(3)$; one has only to be careful about factors of $2$.)
In general, for arbitrary $n \times n$ matrices $X$ and $Y$ a naive application of Cauchy-Schwarz gives $|[X, Y]|^{2} \leq 4|X|^{2}|Y|^{2}$. That the inequality is true with $2$ in place of $4$ goes back to a paper of Chern, do Carmo, and Kobayashi about minimal immersions into spheres, where they proved it for symmetric matrices (it is straightforward to adapt the proof for skew-symmetric matrices). It was shown with $2$ for general matrices by Böttcher and Wenzel (and also by Lu and probably some others too) to hold for any complex matrices, and the inequality in this form is often called the Böttcher-Wenzel inequality. 

For skew-symmetric matrices, the $2$ can be improved to $1$. Lemma 2.5 of [this article][1] of J. Ge shows that, for skew-symmetric $n \times n$ matrices $A, B$, there holds $|[A, B]|^{2} \leq c|A|^{2}|B|^{2}$ with $c = 1/2$ if $n = 3$, and $c = 1$ if $n > 3$. The characterization of the equality case is given too.




  [1]: https://www.sciencedirect.com/science/article/pii/S000187081300385X