Break frequency in ARMA's power spectral density (PSD) Consider first a Lorentzian PSD:
$$L(f) = \frac{\sigma^2}{\alpha^2+(2\pi f)^2}.$$
Its so-called break frequency is at $f_{\rm break} = \frac{\alpha}{2\pi}$, which nicely corresponds to the PSD's shape in a log-log plot (with $\alpha = 1$ below):


The PSD of an ARMA process often has a similar feature. For example, consider the ARMA(1,2) process, with variance $\sigma^2 = 0.05$, AR coefficient 0.89, and MA coefficients $-0.13$ and $-0.07$. Its PSD
$$P(f) = \frac{0.05 [-0.2418 \cos (f)-0.14 \cos (2 f)+1.0218]}{1.7921\, -1.78 \cos (f)}$$

has a similar break, visually at about $f\sim 0.1$.

Q: How can I extract the break frequency (frequencies?) for an arbitrary (stationary) ARMA(p,q) process?
 A: It looks like you may be interested in estimating where the power-law behaviour in the spectrum starts (the break frequency sounds like it would be $f_{min}$ in the below).
Assuming this is correct, the reference shows that you would first assume a model for your spectrum, $S(f)$, to be
$$S(f) = Af^{-\alpha}$$
where $f$ is the frequency, $A$ is a proportionality constant, and $\alpha$ is the scaling parameter of the power law. This model is valid for $f \geq f_{min}$ and so the next step is determining $f_{min}$.
You can estimate $\alpha$ using Eq. (3.1):
$$\hat{\alpha} = 1 + N \sum_{i = 1}^{N} \ln \frac{f_{i}}{f_{min}}$$
The $f_{min}$ value comes from Eq. (3.9):
$$D = \min_{f_{min}} \left( \max_{f \geq f_{min}} |S(f) - P(f)| \right)$$
Where $S(f)$ is your spectrum and $P(f)$ is the power law model using $\hat{\alpha}$ above.
The fact that you have the theoretical spectrum from known ARMA parameters means that perhaps this is not quite what you are looking for and it is highly likely that the paper does a much better job of explaining the process than I have.
Reference:
Clauset, Aaron, Cosma Rohilla Shalizi, and Mark EJ Newman. "Power-law distributions in empirical data." SIAM review 51.4 (2009): 661-703.
A: I've come up with an observation. The function $fL(f)$ has a local maximum at $f_{\rm break}$. Because the ARMA PSDs, $P(f)$, that I'm interested in resemble a Lorentzian (e.g. the one in my question), I'd expect they have a break at the same frequency as the maximum of $fP(f)$. This also exploits the definition of a break frequency by direct analogy to a Lorentzian. For example, the PSD from the question has a break at $f_{\rm break}\approx 0.118861$ (found numerically):


This is not a general approach, though, as the PSD of an ARMA process need not necessarily resemble a Lorentzian, and in some cases (e.g. when the break is at very high frequencies), the function $fP(f)$ might not have a local maximum at the break frequency.
