Is there some sense in which the category $sVect$ of super-vector spaces is the "*maximal non-trivial extension*" of $Vect$ as a symmetric monoidal category?


Is the $\mathbb Z/2$ that shows up in the definition of $sVect$ some kind of homology group of $Vect$?



Is super-vector spaces a "universal central extension" of vector spaces?
This will be a little imprecise, but hopefully if you need a precise result like this you can fill in the details. 


First of all Vect has not only the symmetric monoidal structure but also the direct sum. If you don't look for extensions which are linear for the direct sum, then I think you can form crazy extensions which behave differently depending on the dimension of the vector space. 


But let's assume that your extension is linear and that everything in the extension E is a direct sum of elements in the picard category pic(E) inside your symmetric monoidal category. 


Under these assumptions the extension of symmetric monoidal categories


$$ Vect \to E $$


is completely determined by an extension of symmetric monoidal 2-groups:


$$ pic(Vect) \to pic(E)$$


Now a symmetric monoidal 2-group is classified by two abelian groups: the objects $\pi\_0$, the automorphisms of the identity $\pi\_1$, and a (stable) k-invariant:
$$k : \mathbb{Z}/2 \otimes \pi\_0 \to \pi\_1 $$


This later is the same as a map from $\pi\_0$ into the 2-torsion of $\pi\_1$.


For $pic(vect) = lines$ we have $\pi\_0 = 0$, and $\pi\_1 = K^{\times}$, the group uf units of your ground field. The universal extension then has the same $\pi\_1$, but has $\pi\_0$ equal to the (multiplicative) 2-torsion of $\pi\_1$ with the canonical $k$-invariant. 


If you are not in characteristic 2, this means $\pi\_0 \; pic(E)= \mathbb{Z}/2$, and E is super vector spaces.


If you are in characteristic 2, then sVect is the same as Z/2-graded vector spaces and is no longer universal (the extension splits in the sense you described in your comment). 


