They are not rare. A general construction goes as follows. Start with a finite (étale) group (scheme) $G$ which acts on varieties $F$ and $\tilde Y$, where the second action has no fixed points.
Let $Y =\tilde Y/G$. Then $(F\times \tilde Y)/G\to Y$ has all its fibres isomorphic to $F$. It is locally trivial in the étale topology, but not usually in the Zariski topology. This includes the above examples, and many others.
There is a very general criterion for a map on $\pi_1$ to be surjective. Recall that for $X$ connected, the category of finite étale covers of $X$ is equivalent to the category $\pi_1(X)\text{ -}\operatorname{Set}_f$ of finite sets with a continuous $\pi_1(X)$-action. Under this correspondence, the $Y \to X$ finite étale with $Y$ connected correspond to the connected $\pi_1(X)$-sets $S$ (i.e. $\pi_1(X)$ acts transitively on $S$).
Lemma. Assume $X$, $Y$ connected, and $f \colon X \to Y$ a morphism. Then the induced morphism $\pi_1(f) \colon \pi_1(X) \to \pi_1(Y)$ is surjective if and only if for every $Z \to Y$ finite étale with $Z$ connected, the pullback $Z_X \to X$ is connected.
Proof. If $\pi_1(f)$ is surjective, then clearly any connected $\pi_1(Y)$-set is connected as $\pi_1(X)$-set. Conversely, if $\pi_1(f)$ is not surjective, then some $\gamma \in \pi_1(Y)$ is not in the image. Since fundamental groups are profinite, the image of $\pi_1(f)$ is closed, so the image of $\pi_1(f)$ misses some open neighbourhood of $\gamma$. Thus, there exists an open subgroup $U \subseteq \pi_1(Y)$ such that
$$\gamma U \cap \operatorname{im} \pi_1(f) = \varnothing.$$
Then the finite $\pi_1(Y)$-set $S = \pi_1(Y)/U$ is not connected as $\pi_1(X)$-set. But it is clearly connected as $\pi_1(Y)$-set. $\square$
To apply this to the specific geometric setting you are interested in, just note that if $f \colon X \to Y$ has connected geometric fibres, then the same holds for the base change to any finite étale covering $Z \to Y$. It is then clear that if $Z$ is connected, so is $Z \times_Y X$.
Remark. There are more equivalent criteria for surjectivity; see for example Tag 0B6N. The one I gave above is amongst the ones listed, but this was not the case at the time of writing; hence my writing out the proof. My proof above was originally part of the proof of Tag 0BTX.
Best Answer
I feel like I already answered this question, but it might have been a variant with fibers isomorphic to tori. Let the base $B$ be $\mathbb{A}^2$ with coordinates $s$ and $t$. Begin with $B\times \mathbb{P}^3$, where homogeneous coordinates on $\mathbb{P}^3$ are $[x,y,z,w]$. Let $S$ be the Cartier divisor in $B\times \mathbb{P}^3$ with defining equation $yz-(sx+tw)^2=0$. Let $L$ be the Cartier divisor in $S$ with defining equation $y+z-2(sx+tw)=0$. Let $U$ be the complement of $L$ in $S$. Then $U$ is affine, the morphism $U\to B$ is smooth, and the fiber over every point other than $(0,0)$ is isomorphic to $\mathbb{A}^2$. Of course the fiber over $(0,0)$ is isomorphic to a disjoint union of two copies of $\mathbb{A}^2$. Thus, define $V\subset U$ to be the open subscheme obtained by removing one of these two copies of $\mathbb{A}^2$, i.e., remove the closed subscheme with defining equations $s=t=z=0$. Then $V$ is quasi-affine, and the affine hull is $U$; this follows by Hartog's theorem / the Riemann extension theorem / S2 extension. Therefore $V$ is not isomorphic to an affine space. However, the projection $V\to B$ has all of the requisite properties.
Edit. The older answer I mention above was similar, but a little bit different. That answer was in response to the following similar question, When is a holomorphic submersion with isomorphic fibers locally trivial?.