Let $V$ be an algebraic variety.  If there is a finite ascending chain of Zariski closed sets $\emptyset=V_0\subset V_1\subset \cdots \subset V_n=V$ such that $V_i-V_{i-1}$ is a fintie disjoint union of copies of affine space $\mathbb{A}^i$ we say $V$ is *affine paved* (so $V$ is "algebraically cellular").  

Note: there are non-equivalent variations of this definition (see [here][1]).

One can deduce that an affine paved variety (over $\mathbb{C}$) has no odd cohomology and its even cohomology is free abelian.

Examples:

1. Affine space (or finite unions thereof) is (are) affine paved: $\emptyset\subset \mathbb{A}^n$.
2. Projective space is affine paved since removing a point gives an affine open.  
3. The Bruhat cells in a flag variety show there are non-trivial projective examples.

> Question: Are there non-trivial affine paved affine varieties?

This very well might be a silly question.  Perhaps the affine cone over an affine paved projective variety always works?  It doesn't seem clear to me, and I figured someone out there might have thought about such examples before (Google doesn't seem to know any).

  [1]: https://mathoverflow.net/questions/186832/optimal-definition-of-paving-by-affine-spaces