In my lecture, it says that $Sp(2n, \mathbb{R}) \subset SL(2n, \mathbb{R})$, so the elements of the linear symplectic group have determinant 1.
The explanation is: $\dfrac{1}{n!} \omega_{st}^n = e_1^{*} \wedge f_1^{*} \wedge... \wedge e_n^{*} \wedge f_n^{*}$
and this is up to sign the standard volume form.
I don't see how the claim follows from this. What does the volume form have to do with the determinant?
If $f\in Sp(2n)$, $f^*\omega_{st}=\omega_{st}$ implies that $f^*\omega_{st}^n=(f^*\omega_{st})^n=\omega_{st}^n$. This implies that $f$ preserves the volume form $\omega=\omega_{st}^n$. Remark that $f^*\omega=det(f)\omega$, we deduce that $f\in Sl(2n,\mathbb{R})$.