In terms of vector field analogies to closed and exact differential forms, conservative and incompressible vector fields (gradient and divergence) generalize to higher dimensions, but curl and irrotational fields do not. Why? 
Cross product doesn't generalize either but one can use exterior products and hodge duals to fullfill the need. In differential geometry, there is a duality between the boundary operator on chains and the exterior derivative as expressed by the general Stokes' theorem. 
By a theorem of De Rham, the exterior derivative is the dual of the boundary map on singular simplices. From this perspective, there should be a generalized curl. Perhaps, there is an explanation in terms of the Poincare Lemma where for n>3 (but perhaps not 7), curl fails for higher dimensions? 



Generalization of Curl to higher dimensions
Let ${\bf K}$ be a vector field in the neighbourhood of ${\bf p}\in{\mathbb R}^n$, and let ${\bf X}$ and ${\bf Y}$ be two tangent vectors at ${\bf p}$. These two vectors span a parallelogram $P$ with one vertex at ${\bf p}$. The "circulation" of ${\bf K}$ around $P$ computes to
$$
\int\_{\partial P}{\bf K}\cdot \mathrm{d}{\bf x}= (L\,{\bf X})\cdot{\bf Y}- (L\,{\bf Y})\cdot{\bf X} + o(|P|^2)
$$
with $L:=\mathrm{d}{\bf K}({\bf p})$ and $|P|:= \mathrm{diam}(P)$. It follows that there is a certain skew bilinear function ${\rm Rot}{\bf K}({\bf p}):T\_{\bf p}\times T\_{\bf p}\to{\mathbb R}$ with
$$
\int\_{\partial P}{\bf K}\cdot \mathrm{d}{\bf x}={\rm Rot}{\bf K}({\bf p})({\bf X},{\bf Y})+ o(|P|^2) \quad (|P|\to 0).
$$
In the case $n=3$ the bilinear form ${\rm Rot}$ can be represented by the vector ${\rm curl}{\bf K}$ in the form
$$ 
{\rm Rot}{\bf K}({\bf p})({\bf X},{\bf Y}) = {\rm curl}{\bf K}({\bf p})\cdot({\bf X}\times{\bf Y}).
$$


