Define the canonical symplectic form $\omega$ on $\mathbb{R}^{2n}$ by $\omega(u,v)=u^TJv$, where $$J=\begin{bmatrix} 0 & I_n \\ -I_n & 0 \end{bmatrix}.$$
I do not understand why the volume $V(v_1,\ldots,v_n)$ of the parallelepiped spanned by vectors $v_1,\ldots,v_n$ in $\mathbb{R}^{2n}$ is given by $$\det (\omega(v_i,v_j)).$$ I understand that $ω^n$ is a volume form. My feeling is that I must show that $$ \omega^n(v_1,\ldots,v_{2n})=\det (\omega(v_i,v_j)). $$ But I do not have a proof of this equality.
I assume you mean $v_1, \dots, v_{2n}$ rather than $v_1, \dots, v_{n}$.
The volume of the parallelepiped spanned by $v_1, \dots, v_{2n}$ is just $V(v_1, \dots, v_{2n}) = \left|\det(v_1, \dots, v_{2n})\right|$, right?
So what about $\det\left(\omega(v_i, v_j)_{1 \leqslant i,j\leqslant 2n}\right)$? Well, let's call $A$ this matrix $A = \left(\omega(v_i, v_j)_{1 \leqslant i,j\leqslant 2n}\right)$. This is just the matrix associated to $\omega$ is the basis $v_1, \dots, v_{2n}$ (I'm assuming here that $(v_1, \dots, v_{2n})$ are linearly independent, otherwise the result is trivial). So we can write $A = P^T\,J\,P$, where $P$ is the change-of-coordinates matrix (from the canonical basis of $\mathbb{R}^{2n}$ to the basis $(v_1, \dots, v_{2n})$). Therefore $\det(A) = \det(P)^2 \det(J)$. Now, $\det(P) = \det(v_1, \dots, v_{2n})$ (by definition, almost) and you can check that $\det(J) = 1$. In conclusion, I believe the correct result is: $$\det\left(\omega(v_i, v_j)_{1 \leqslant i,j\leqslant 2n}\right) = \left[V(v_1, \dots, v_{2n})\right]^2$$
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NB: About your idea "$\omega^n$ is a volume form": indeed, and you can check that this volume form is $\omega^n = \lambda \det$, where $\lambda = \det(J)$, in this case $\lambda = 1$. In other words here $\omega^n$ is the standard volume form on $\mathbb{R}^{2n}$ (i.e. $\omega^n = \det$). This means in particular that $V(v_1 \dots, v_{2n}) = \left|\omega^n(v_1, \dots, v_{2n})\right|$. Maybe you can derive the previous identity from this observation.