= Solution
At late time the Gaussian curvature <power spectrum> is $P(k)=H^2/(4\epsilon M_{\mathrm{Pl}}^2k^3)$. Consider the <squeezed bispectrum configuration> $k_1=q\ll k_2\simeq k_3=k$. The <local-type primordial non-Gaussianity> result is dominated by
$$
B^{\mathrm{loc}}(q,k,k)\simeq\frac{12}{5}f_{\mathrm{NL}}P(q)P(k).
$$
In the <curvature bispectrum from a zeta zeta-prime-squared interaction>, the leading cyclic term has the long mode in the undifferentiated slot. Its momentum numerator tends to $k^4/(2k)=k^3/2$, while the other two cyclic terms contain $q^2$. Hence
$$
\boxed{B^{\mathrm{sf}}(q,k,k)\simeq\frac{H^4}{32\epsilon M_{\mathrm{Pl}}^4q^3k^3}=\frac\epsilon2P(q)P(k).}
$$
This is the <slow-roll amplitude of the squeezed zeta zeta-prime-squared bispectrum>. It retains the local-like $q^{-3}$ enhancement: assigning the soft momentum only to differentiated legs would incorrectly miss the dominant permutation. Comparing squeezed amplitudes gives
$$
\boxed{f_{\mathrm{NL}}^{\mathrm{effective}}=\frac{5\epsilon}{24}\ll1.}
$$
For comparison, in the equilateral configuration $B^{\mathrm{sf}}(k,k,k)=(4\epsilon/3)P(k)^2$, so the signal is slow-roll suppressed away from the squeezed limit as well.
Use the quoted observational bound as the sensitivity benchmark supplied by this problem, rather than as a new current measurement. A local-template sensitivity at the order-ten level is far above this order-$\epsilon$ signal; detection of this interaction with comparable <CMB> measurements is consequently very unlikely. The entire shape is not identical to the local template, so its bound is not an exact constraint on every single-field shape, but the squeezed amplitude already exposes the large suppression. \b[The obstacle is the slow-roll amplitude, even though this specified interaction has a squeezed enhancement.]
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