Integrate over a large semi-circle with a tiny indentation around $0$. The integral of $f$ along every such contour is $0$, and the part coming from the large semi-circle will tend to $0$ as the radius tends to $+\infty$.
Finally the integral over the tiny indentation will be $0$ for every $\varepsilon > 0$, either by parametrizing or by using a primitive for $f$ (and this is despite $f$ having a triple pole at the origin.) But as you say, you won't be able to show this by just estimating $|f|$ on the indentation.
Integrals of the form
$$\int_{-\infty}^\infty \frac{p(x)}{\cosh x}\,dx,$$
where $p$ is a polynomial can be evaluated by shifting the contour of integration to a line $\operatorname{Im} z \equiv c$. We first check that the integrals over the vertical segments connecting the two lines tend to $0$ as the real part tends to $\pm\infty$:
$$\lvert \cosh (x+iy)\rvert^2 = \lvert \cosh x\cos y + i \sinh x\sin y\rvert^2 = \sinh^2 x + \cos^2 y,$$
so the integrand decays exponentially and
$$\left\lvert \int_{R}^{R + ic} \frac{p(z)}{\cosh z}\,dz\right\rvert
\leqslant \frac{K\,c}{\sinh R}\left(R^2+c^2\right)^{\deg p/2} \xrightarrow{R\to \pm\infty} 0.$$
Since $\cosh \left(z+\pi i\right) = -\cosh z$, and the only singularity of the integrand between $\mathbb{R}$ and $\mathbb{R}+\pi i$ is a simple pole at $\frac{\pi i}{2}$ (unless $p$ has a zero there, but then we can regard it as a simple pole with residue $0$) with the residue
$$\operatorname{Res}\left(\frac{p(z)}{\cosh z};\, \frac{\pi i}{2}\right) = \frac{p\left(\frac{\pi i}{2}\right)}{\cosh' \frac{\pi i}{2}} = \frac{p\left(\frac{\pi i}{2}\right)}{\sinh \frac{\pi i}{2}} = \frac{p\left(\frac{\pi i}{2}\right)}{i},$$
the residue theorem yields
$$\begin{align}
\int_{-\infty}^\infty \frac{p(x)}{\cosh x}\,dx
&= 2\pi\, p\left(\frac{\pi i}{2}\right) + \int_{\pi i-\infty}^{\pi i+\infty} \frac{p(z)}{\cosh z}\,dz\\
&= 2\pi\, p\left(\frac{\pi i}{2}\right) - \int_{-\infty}^\infty \frac{p(x+\pi i)}{\cosh x}\,dx\\
&= 2\pi\, p\left(\frac{\pi i}{2}\right) - \sum_{k=0}^{\deg p} \frac{(\pi i)^k}{k!}\int_{-\infty}^\infty \frac{p^{(k)}(x)}{\cosh x}\,dx.\tag{1}
\end{align}$$
Since $\cosh$ is even, only even powers of $x$ contribute to the integrals, hence we can from the beginning assume that $p$ is an even polynomial, and need only consider the derivatives of even order.
For a constant polynomial, $(1)$ yields
$$\int_{-\infty}^\infty \frac{dx}{\cosh x} = 2\pi - \int_{-\infty}^\infty \frac{dx}{\cosh x}\Rightarrow \int_{-\infty}^\infty \frac{dx}{\cosh x} = \pi.$$
For $p(z) = z^2$, we obtain
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx &= 2\pi \left(\frac{\pi i}{2}\right)^2 - \int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx - (\pi i)^2\int_{-\infty}^\infty \frac{dx}{\cosh x}\\
&= - \frac{\pi^3}{2} - \int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx + \pi^3,
\end{align}$$
which becomes
$$\int_{-\infty}^\infty \frac{x^2}{\cosh x}\,dx = \frac{\pi^3}{4}.$$
Best Answer
The Cauchy Principal Value of the integral of interest is given by
$$\begin{align} \text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)&=\lim_{\varepsilon\to 0^+}\left(\int_{-\infty}^{-1-\varepsilon} \frac{e^{ipx}}{x^4-1}\,dx\int_{-1+\varepsilon}^{1-\varepsilon} \frac{e^{ipx}}{x^4-1}\,dx\int_{1+\varepsilon}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right) \end{align}$$
We shall analyze the case for which $p>0$.
METHODOLOGY $1$:
Now, take $R>1$. If we evaluate the contour integral $\displaystyle \oint_C \frac{e^{ipz}}{z^4-1}\,dz$ where the contour $C$ is comprised of $(i)$ the real line segments from $-R$ to $-1-\varepsilon$, $(ii)$ the semi-circular arc in the third quadrant centered at $-1$ with radius $\varepsilon$ from $-1-\varepsilon$ to $-1+\varepsilon$, $(iii)$ the straight line segment from $-1+\varepsilon$ to $1-\varepsilon$, $(iv)$ the semi-circular arc in the first quadrant centered at $1$ with radius $\varepsilon$ from $1-\varepsilon$ to $1+\varepsilon$, $(v)$ a straight line segment from $1+\varepsilon$ to $R$, and $(vi)$ a semicircular arc from $R$ to $-R$, then the Residue theorem guarantees that
$$\oint_C \frac{e^{ipz}}{z^4-1}\,dz=2\pi i \text{Res}\left(\frac{e^{ipz}}{z^4-1}\,dz, z=i\right)=-\frac{\pi}{2}e^{-p}$$
As $R\to \infty$ and $\varepsilon\to 0^+$, we see that
$$\lim_{R\to\infty\\\varepsilon\to 0^+}\oint_C \frac{e^{ipz}}{z^4-1}\,dz=\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)+\frac\pi2\sin(p)$$
Putting it together, we find that
$$\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)=-\frac\pi2\left(\sin(p)+e^{-p}\right)$$
METHODOLOGY $2$:
Using partial fraction expansion, we can write
$$\frac{e^{ipx}}{x^4-1}=\frac{e^{ip}}4 \frac{e^{ip(x-1)}}{x-1}-\frac{e^{-ip}}4 \frac{e^{ip(x+1)}}{x+1}+\frac{ie^{-p}}4 \frac{e^{ip(x-i)}}{x-i}-\frac{ie^{p}}4 \frac{e^{ip(x+i)}}{x+i}$$
Then, we have
$$\begin{align} \text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)&=\frac{e^{ip}}4 \text{PV}\left(\int_{-\infty}^\infty \frac{e^{ip(x-1)}}{x-1}\,dx\right)\\\\ &-\frac{e^{-ip}}4\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ip(x+1)}}{x+1}\,dx\right)\\\\ &+\frac{ie^{-p}}4\int_{-\infty}^\infty \frac{e^{ip(x-i)}}{x-i}\,dx\\\\ &-\frac{ie^{p}}4\int_{-\infty}^\infty \frac{e^{ip(x+i)}}{x+i}\,dx\tag1 \end{align}$$
The Cauchy Principal values of the first two integrals on the right-hand side of $(1)$ are identical and equal to the value of the integral $\displaystyle \int_{-\infty}^\infty \frac{\sin(px)}{x}\,dx=i\pi\text{sgn}(p)$. For $p>0$ ($p<0$), the Residue Theorem guarantees that the value of the fourth (third) integral in $(4)$ is $0$, while the value of the third (fourth) integral is $2\pi i$ ($-2\pi i$).
Putting it together, we find that
$$\text{PV}\left(\int_{-\infty}^\infty \frac{e^{ipx}}{x^4-1}\,dx\right)=-\frac\pi2 \left(\sin(|p|)+e^{-|p|}\right)$$
as expected!