Is there a finite unitary gate set which can exactly realise all QFTs of order $2^n$?
No, there is no decomposition of the entire family $\{F_{2^n}\}_{n\geqslant1}$ into a single finite gate-set. Here's why.

The QFTs involve only coefficients over $\overline{\mathbb Q}$, the complex algebraic closure of the rational numbers. In analogy to [Adleman+Demarrais+Huang&ndash;1997], if we involved any gates which included any transcendental numbers, we could choose a minimal set of transcendentals $\{\tau_1, \tau_2, \ldots\}$ and describe the gate-coefficients essentially as rational functions $\overline{\mathbb Q}(\tau_1, \tau_2, \ldots)$. To obtain the QFT as a product of such gates, we must arrange for all of the transcendental components to cancel (a similar thing must occur to ensure each of the gates are unitary); but then we might as well replace all of the transcendentals with $0$, so that all of the coefficients are algebraic. So we restrict ourselves to algebraic gate-sets without loss of generality.

The coefficients of a finite gate set over $\overline{\mathbb Q}$ can all be contained in a finite-degree extension of $\mathbb Q$, which one may construct by extending $\mathbb Q$ by those very coefficients. However, the gates $\mathrm{CZ}_{2^n}$ obviously have coefficients belonging to field extensions over $\mathbb Q$ of degree $2^{n-1}$, i.e.&nbsp;of unbounded degree.
Thus the family of QFTs of order $2^n$ does not decompose into any finite gate-set.

As a corollary, we cannot hope to have any algorithms in $\mathsf{EQP}$ which relies on QFTs over cyclic rings of unbounded size &mdash; note that the same problem occurs for any family of circuits which might use QFTs of arbitrary order.