Simple computation with the $n$-form $dz_1\wedge...\wedge dz_n$ in $\mathbb{C}^n$

You're right. But $dz_1\wedge\dots\wedge dz_n (e_1,\dots,e_n) = 1$. Note, for example, that in $\Bbb C$, we have $dz(e_1) = (dx+i\,dy)(e_1) = dx(e_1)+i\,dy(e_1)=1+i\cdot 0 = 1$. Extrapolating, $dz_1\wedge\dots\wedge dz_n(e_1,\dots,e_n) = dz_1(e_1)dz_2(e_2)\cdots dz_n(e_n) = 1$, since $dz_k(e_j)=0$ when $k\ne j$.

If you have a n-dimensional manifold, then any n-form is a multiple of the det function, since the n-forms are a 1-dimensional space and det is an alternating function. If you take the dual standard basis, the n-form $dz_1 \wedge \ldots \wedge dz_n(v_1\ldots,v_n)$ is the det of the matrix whose columns are $v_1,\ldots v_n$.