Also want to point out any generalization of curl to 4 dimensions leads to a 6 dimensional value. But I'll first explain 3 dimensional curl.
Three dimensional curl can be geometrically interpreted as a rate of rotation induced by a vector field. It can be described using a 3 by 3 matrix representing the "induced rotation" in the plane generated by each pair of axis (x, y and z). However, this matrix must be anti-symmetric. A pair of like axis doesn't generate a plane and thus the diagonal elements (x-x, y-y, and z-z) aren't used. Also any rotation in the the x-y plane is just the negative of the same rotation in the y-x plane (same goes for y-z vs z-y and z-x vs x-z). Therefore there are really only three planes needed to completely describe a rotation in 3-space, namely x-y, y-z, and z-x.
It turns out in three dimensions each plane has exactly one normal vector, so each plane rotation can be described as a rotation about a given axis using the right hand rule. However this doesn't hold in 4 dimensions because there is no longer a single normal vector for each plane. Instead there is an entire plane of vectors perpendicular to any given plane. Therefore the common notion of "rotation about an axis" no longer makes any sense. Vector fields in dimensions other than 3 do not have a vector form of curl.
Instead the rotation must be described from an anti-symmetric 4 by 4 matrix, which turns out to have 6 linearly independent elements. In the notation of differential forms the curl of a 4 dimensional vector field is not a vector but a bi-vector. The 6 differential rotation components are 2 dimensional objects (dx^dy, dy^dz, dz^dw, dw^dx, dx^dz, and dy^dw).
So in general, the n-dimensional "curl" will be bi-vector with n(n-1)/2 linearly independent components. It just so happens that when n is 3 we get 3*(3-1)/2 = 3 so there is an isomorphism between the bi-vector form and the vector form. Vector fields in \mathbb{R}^n with n something other than 3 do not have a vector form curl, but all have a bi-vector form curl.