What numbers can be created by repetitive division of subsequent natural numbers?
The answer is no you cannot get every rational number. For example you cannot get the number $2$. I'll need to use Bertrand's postulate to explain why:
For every $n>1$ there must be a prime $p$ such that $n < p < 2n$. Or written slightly differently, for any $n > 2$ there must be a prime $p$ such that $\lfloor n/2 \rfloor < p < n$.
Suppose you have a tower equation of size $n$. You can resolve all the parenthesis in your tower equation and write it as a proper fraction
$$\frac{k_1 \dotsb k_\ell}{k_{\ell+1} \dotsb k_n}\,.$$
where the $k_i \in \{1, \dotsc, n\}$ and $k_i = k_j \implies i = j$, so the $k_i$ are just some ordering of $\{1, \dotsc, n\}$. One of these values of $k_i$ must be a prime $p$ between $\lfloor n/2 \rfloor$ and $n$ such that $p$ does not divide any of the other numbers $\{1, \dotsc, n\}$ besides itself. So in your proper faction, there will be exactly one factor of $p$ in the numerator or denominator. If your size $n$ is at least four, this prime isn't $2$, and since we can write out all the values of the tower equations of size four or less (see TheSimpliFire's answer) and see that the value of $2$ doesn't appear, $2$ cannot appear for a larger size tower because of this "un-cancellable" prime $p \neq 2$ that must appear in either the numerator or denominator.
Here's an unrelated thought I had about this question. I don't immediately see how it can help answer the question, but it does mean that the number of tower equations of size $n$ will be the $n^{\text{th}}$ Catalan number.
Think about building a tower as writing down your large fraction without parenthesis, and then choosing the "top-most" fraction bar, dividing your tower into two smaller towers, the numerator and denominator, and then continuing this process recursively on those two smaller towers. Doing this on a tower of size $n$ builds a full binary tree with $n$ leaves, the numerators corresponding to left branches and the denominators corresponding to right branches, and with the leaves labelled in order from left to right with the numbers $\{1, \dotsc, n\}$. Then the number that the tower equals can be calcated from the tree as
$$\prod_{i\in\{1, \dotsc, n\}} (-1)^{r_i}i$$
where $r_i$ is the number of "right turns" or "right branches" in the tree you'd have to take to get to the leaf $i$.