Two questions on the Grothendieck ring of varieties 1) In the definition of the Grothendieck ring of varieties over a field $k$, which definition of the various notions of "variety" is chosen? Finite type and separated, or maybe more?
2) If $\mathbb{L}$ is the class of the affine line in the Grothendieck ring, does the class of $\mathrm{GL}_n$ equal $(\mathbb{L}^n - 1) \cdot \dotsc \cdot (\mathbb{L}^n - \mathbb{L}^{n-1})$? Somehow this should be true, but for this I would need a "fibration relation", which does not seem to follow from the scissor relation.
 A: A little comment which does not seem to involve algebraic spaces. 
All I am saying here I learnt in this paper.
Basically, given any two Zariski fibrations $\pi_i:X_i\to Y$, $i=1,2$, with the same fiber $F$, one has the relation $[X_1]=[X_2]$ in the Grothendieck ring. Indeed, one can stratify the base $Y=\coprod_j U_j$ (these $U_j$'s being connected locally closed subvarieties) so that $\pi_i:{\pi_i^{-1}(U_j)\to U_j}$ are trivial, thus $$[X_1]=\sum_j[\pi_1^{-1}(U_j)]=\sum_j([U_j]\cdot [F])=\sum_j[\pi_2^{-1}(U_j)]=[X_2].$$ Both classes equal $[Y]\cdot [F]$.

The fact that $$[\textrm{GL}_n]=\prod_{i=0}^{n-1}(\mathbb L^n-\mathbb L^i)$$ can be proved by induction. For $n=1$ it is clear. For arbitrary $n$, use the fibration $\pi:\textrm{GL}_n\to \mathbb C^n\setminus 0$ sending a matrix to its first column, say. The fiber is $\textrm{GL}_{n-1}\times \mathbb C^{n-1}$, so you can write $[\textrm{GL}_n]=(\mathbb L^n-1)\cdot \mathbb L^{n-1}\cdot [\textrm{GL}_{n-1}]$, and you can conclude by induction. 
If you work with Grothendieck rings, it might be interesting to know that the classes $[\textrm{GL}_n]$ are exactly the classes you need to invert in $K_0(\textrm{Var}_k)$ in order to get the Grothendieck ring of stacks:
$$K_0(\textrm{Var}_k)[[\textrm{GL}_n]^{-1}\,|\,n\geq 1]\cong K_0(\textrm{St}_k).$$

For your question (1), unfortunately I do not know enough literature to guess what is the most common definition of variety. But reduced seems redundant to me. Indeed, the closed immersion $X_{\textrm{red}}\to X$ has empty open complement, and the scissor relation then tells you $[X]=[X_{\textrm{red}}]$.
A: From what I've read, it seems like separated and finite type are fairly standard. Poonen, in his paper "The Grothendieck Ring of Varieties is not a Domain" adds geometrically reduced- I haven't seen this other places, but I haven't looked too hard.
For the second, there does exist a fibration relation- seeing it is difficult without going to more complicated categories with isomorphic Grothendieck rings. Consider the inclusion $\mathrm{Var}_k\to\mathrm{Space}_k$- this gives rise to a homomorphism of grothendieck rings, which turns out to be an isomorphism. The following argument then works in the category of algebraic spaces over $k$, and the relation in fact holds in the Grothendieck ring of varieties:
Consider a fibration $E\to B$ with fiber $F$. In the category of algebraic spaces, there exists an open set $U\subset B$ such that $E_U\to U$ is isomorphic to $U\times F\to U$. Let  $V= B\setminus U$, a closed subset of $B$. So we have $[E]=[F][U]+[E_V]$, and by noetherian induction, we may say that $[E_V]=[F][V]$, and so $[E]=[F][B]$. Now you can apply the argument you seem to have in mind in (2).
