Computing the k-volume of a k-parallelogram (i.e. a parallelogram spanned by k n-dimensional vectors) in n dimensions is straightforward: Let $P=[\overrightarrow{v_1},...,\overrightarrow{v_k}]$, then the k-volume is equal to $\sqrt{det(P^{T}P)}$. For instance, if n=3 and k=2, then this formula reduces to the norm of the cross-product and thus gives the 2-volume (area) of the parallelogram spanned by two 3-D vectors.
Now this fact is very useful when trying to find the k-volumes of manifolds (e.g. areas of 2-D surfaces in 3-D) by integration.
My question is, is there differential form or a form field that would serve the same purpose? In $\mathbb{R}^3$ for instance, the elementary 2-forms give you the signed area of the "projection" of the 2-parallelogram onto (x,y) or (x,z) or (y,z) planes depending on the form you use, but not the area of 2-parallelogram living in the space.
No, you need to know what plane $\vec n\cdot\vec x=0$ your parallelogram in $\mathbb R^3$ lies in to write down the $2$-form that gives oriented area of the parallelogram. One of the distinguishing features of complex (differential) geometry is that there is, for each $k=1,\dots,n$, a $2k$-form on $\mathbb C^n$ that gives the $(2k)$-dimensional area form on every $k$-dimensional complex subspace.