Do linear momentum eigenstates exist? Suppose that a particle  is in  a  linear momentum eigenstate $\mid p \rangle$ that is $\hat p \mid p \rangle=p \mid p \rangle$. According to the Heisenberg uncertainty principle we must have  $p\neq 0$. I am not familiar with how detectors work, but they should have finite size. So if a  detector detects a particle with zero momentum then the particle should remain in the detector. To not contradict the Heisenberg uncertainty principle the detector should have infinite size.
Now there is a reference frame where this particle has linear momentum   equal to zero so in  this frame the particle is represented by  $\mid 0 \rangle $.
How can we resolve this apparently contradiction?
 A: $\require{cancel}$
Well, they "exist" in our (physicists') imagination, as a limit, but they are not normalizable.
Adopting your notation for the null eigenstate of the momentum operator, $\hat p|0\rangle=0$, you may readily identify it as Dirac's celebrated standard ket, through which he defines his bra-ket formalism in Ch III.20 of his textbook. But, since $\langle p|0\rangle=\delta(p)$, your hyper-localized state has infinite norm, "$\delta(0)$".  You may, if you wished, approximate it as the limit of a suitably normalized Gaussian in momentum space, as its width goes to zero. (Intelligent squeezed coherent states. )

*

*Note it is translationally invariant, since the exponential of the momentum operator acting on it amounts to one.

The x-space wavefunction of it is a plane-wave of zero momentum, so, then, a constant, $\langle x|0\rangle= 1/\sqrt{2\pi \hbar}$, everywhere in space; also unnormalizable, and terminally delocalized: a flatline, translationally invariant. In your Gaussian approximation, it would be a "Gaussian" with infinite width.
But you may translate this state in momentum space by acting on it with the exponential of the position operator,
$$ \exp (ip\hat{x}/\hbar ) | 0\rangle = |p\rangle  ,$$
and further define, with Dirac,
$$|x\rangle= \delta(\hat{x}-x) | 0\rangle  \sqrt{2\pi \hbar} .$$
The uncertainty principle is fine as a freak limit, but you'd better reassure yourself with the Gaussians involving mutually inverse widths. As it stands, it is egregious nonsense, which should never have been written down,$\cancel{
 \frac{\langle 0|\hat p^2 |0\rangle }{\langle 0|0\rangle } \frac{\int dx ~x^2 }{\int dx }  }$. I didn't write this...
