Here is just a compilation of my comments above plus some further comments.

As mentioned in the question, [a subsequence of the sequence of] balls form a natural Folner sequence in any group of subexponential growth. Now, as pointed out by others, balls (w.r.t. to some finite generating set) are fairly "ugly". This can be made precise if one considers the concept of an optimal Folner set:

Let $I(n)= \displaystyle \inf_{|A| \leq n} \dfrac{|\partial A|}{|A|}$ (the $\inf$ runs over all sets $A$ of size $\leq n$) be the isoperimetric profile. 
Then a set $F$ is optimal if $I(|F|)=\dfrac{|\partial F|}{|F|}$. 
In words: if a set $E$ is not larger [cardinality-wise] than $F$, then it's isoperimetric ratio $\dfrac{|\partial E|}{|E|}$, does not beat the isoperimetric ratio of $F$.

One can check (using the Loomis-Whitney inequality) that optimal Folner sets in $\mathbb{Z}^d$ (w.r.t. the usual generating set) are [hyper]cubes (or that they tend to have a rectangular form). 
This is an unambiguous way of saying that balls are not the "nicest" Folner sets. Also optimal sets are not "weird" at all (since they must be extremely well-chosen). 

Next, given a group of exponential growth, it's an open question whether any subsequence of the sequence of balls is Folner. I gave a [partial answer](https://mathoverflow.net/questions/148995/folner-sets-and-balls/235348#235348) which shows this is not case when the group [together with the choice of generating set] has pinched exponential growth. This includes many wreath products, solvable Baumslag-Solitar groups and some extensions of $\mathbb{Z}^d$ by $\mathbb{Z}$. 

These groups can all be written as semi-direct products. 
If $G$ and $H$ are amenable, then one can show that $G \rtimes H$ is amenable and that Folner sets are of the Form $E_n \times F_n$ (where $E_n$ [resp. $F_n$] is a Folner sequence of $G$ [resp. $H$]). 
In that sense, the Folner sets that we come across (lazily, in the sense that they are produced by a general proof) in such groups are "rectangular". 

Hence the groups mentioned above [that solvable Baumslag-Solitar, some metabelian groups, any group whose growth series do not have a double pole at the radius of convergence (which includes many wreath products)] are the most direct answers to your question (at least for some generating set). One knows that balls (w.r.t. generating sets are not Folner) but some "rectangular" set is (just to be precise: there could be groups with a single pole which are not semi-direct products or extensions of amenable groups; for these groups [if any are known] there are no "rectangular" sets).

For non-split extensions a description of the Folner sets was [given over there](https://math.stackexchange.com/questions/2928075/extensions-of-amenable-groups).
Note one could adapt the meaning of "rectangular" for non-split extensions: by taking a preimage of the Folner set of the quotient times some Folner set of the subgroup.

So now one might think that "rectangular" (and no longer balls) sets are favourites. But then there are also simple groups of intermediate growth [see this question](https://mathoverflow.net/questions/122199/is-it-true-that-every-f-g-infinite-simple-group-has-exponential-growth). And (if not for such groups, then for other simple groups of subexponential growth) I guess that balls are the only candidates one has.

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**Side Note 1:** it's an long-standing open question to prove what are such sets in the (continuous) Heisenberg group (although the conjectured shape is well-described). That was my motivation for [this](https://mathoverflow.net/questions/218460/sharp-isoperimetry-in-the-discrete-heisenberg-group) question.

**Side Note 2:** As pointed out by Ycor, given a Folner sequence $F_n$ you can make it "as weird as you want" by considering an arbitrary sequence of finite sets $E_n$ with $\dfrac{|E_n|}{|F_n|} \to 0$. One the advantage of considering optimal Folner sequences would be to avoid such set-ups (the obvious disadvantage, is that there are almost no groups where optimal sets are known). A further note is that adding such a set $E_n$ has no influence on the invariant measure one obtains (for a fixed ultrafilter). Note that translating the sets can have an effect on the limit measure.