This is not a real answer since there is a gap, a claim I'm not currently able to prove, but it is too long for a comment and I hope somebody can fix the gap. Consider the topological property of being [sequential][1]. The idea is that $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ is sequential but $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$ should not. 

First countable spaces, such as $\mathbb Q\times \mathbb Q$, are sequential. Sequential spaces are closed under quotients and disjoint unions, hence $\mathbb Q\wedge \mathbb Q$, $(\mathbb Q\wedge \mathbb Q)\times\mathbb N$ and $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ are sequential.

We can identify the underlying sets of $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ and $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$ in the obvious way, so we're really speaking about different topologies on the same set. May and Sigurdsson 
show that the former has strictly more open sets than the latter. *I think both of them have the same convergent sequences, but I haven't been able to prove it (the only problem is when the limit is the base point).* If this is true, any open set in $(\mathbb Q\wedge \mathbb Q)\wedge\mathbb N$ would be sequentially open in $\mathbb Q\wedge (\mathbb Q\wedge\mathbb N)$, so the latter cannot be sequential, otherwise the two topologies would be the same.


  [1]: https://en.wikipedia.org/wiki/Sequential_space