$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$
Consider a string of length $L$ under tension and clamped on each end.
This system is described by the wave equation and has a set of modes.
The $n^\text{th}$ mode has a spatial profile
$$\phi_n(x) = \sin(n \pi x / L)$$
and frequency $\omega_n$.
In other words, if only the $n^\text{th}$ mode is excited, the time dependent displacement of the string is
$$y(x,t) = A \cos(\omega_n t) \sin(n \pi x / L)$$
for some amplitude $A$.$^{[a]}$
In fact, the dynamics of each mode is exactly that of a harmonic oscillator with frequency $\omega_n$.
We also treat the string mode by mode when we do quantum mechanics.
Each mode can have an integer number of excitations.
If the $n^\text{th}$ mode has $m$ excitations, we say it is in "$m^\text{th}$ Fock state" and denote it $\ket{m_n}$.
This is the so-called "second quantized" language where the state of the string is written as
$$\ket{m_0}\otimes \ket{m_1} \otimes \ket{m_2} \otimes \ldots = \ket{m_0 m_1 m_2 \ldots} \, .$$
Each Fock state has an associated wave function.
For example, if a mode is in $\ket{0}$, then that mode has a Gaussian probability distribution its quadratures.
This Guassian wave function is completely different from the spatial profile of the mode $\phi_n(x)$.
Now consider an infinite square well.
This system has various eigenstates which we can label $\ket{\text{I}}$, $\ket{\text{II}}$, etc.
If we have only a single particle we usually just describe the system by saying which eigenstate, or superposition of eigenstates, the particle is in.
Each of these eignestates has an associated wave function, i.e. $\langle x|\text{III}\rangle = \phi_\text{III}(x)$.
These wave functions tell us, among other things, the probability distribution of a single particle's position.
With multiple particles, we originally learn to express the total state by specifying which state each particle is in.
For example, if the first particle is in $\ket{\text{I}}$ and the second is in $\ket{\text{III}}$, we'd write
$$\ket{\Psi} = \ket{\text{I}} \otimes \ket{\text{III}} \, .$$
Of course, since particles of a single type are indistinguishable we have to symmetrize or antisymmetrize this state.
It's easier though, to just use second quantization and write
$$\ket{\Psi} = \ket{101}$$
which means there's one excitation in the first state and one excitation in the third state.
Observe that now what we called "eigenstates" for the single particle in the square well seem to play the role of what we called "modes" in the case of the quantized vibrating string.
Perhaps this is not too surprising: we can imagine that there is a quantum field inside the square well and that field has modes just like a vibrating string.
In this picture, what we normally think of as the the single particle wave functions are like the vibrational modes of the string.
In this picture it's natural to say that each mode can be excited by one, two, three… quanta, which is the same as saying that one, two, three… particles can be in each single particle state.
In other words, each mode can be in any of the various Fock states.
However, if we take this perspective, then we expect that each mode's Fock states should have associated wave functions.$^{[b]}$
For example, if we have the state $\ket{0000\ldots}$ in the square well, then we're saying that each mode of the square well has an amplitude which is Gaussian distributed.
In the case of the string, each mode having a ground state with Gaussian distributed amplitude means that if you measure the displacement of the string you don't always get zero.
This is the so-called "zero point motion".
What does it mean for a particular mode of particles in an infinite square well to have a Fock state $\ket{0}$ with Gaussian wave function?
Naively this means that even with no particles in the well, there is nonzero probability of measuring a nonzero displacement in the particle's quantum field.
What does this mean?
$[a]$: We're ignoring phase here.
$[b]$: It's becoming clear to me that the term "state" is overloaded. What we call "single particle states" should be called "modes", and the term "state" should be reserved for the various Fock states of each mode, or for the total system state which is a superposition of tensor products of Fock states.
Best Answer
As far as the many-particle setting is concerned, I am tempted to say that the short answer is "Nothing, really, because the $|0\rangle$ state is not Gaussian." :D
Longer answer: The formal vacuum state of second quantization has nothing to do with the ground state of a harmonic mode in this case. It is rather an abstract state enabling the isomorphism between the (anti)symmetric subspace of the actual N particle Hilbert space and an abstract Hilbert space constructed as a direct product of "particle" or "mode" spaces, each equipped with an individual ladder operator algebra.
One fast way to argue that no vacuum Gaussian wavefunction factors into the construct comes close to validating @MarkMitchison's comment: just check the regular localization probabilities, or even the space correlation functions, on the vacuum state. Say the single-particle eigenfunctions are $\phi_n(x)$ (including the actual ground state!) and the corresponding ladder operators are ${\hat a}_n$, ${\hat a}^\dagger_n$, such that the field operators read $$ {\hat \psi}^\dagger(x) = \sum_n{\phi^*_n(x){\hat a}^\dagger_n}, \;\;\;{\hat \psi}(x) = \sum_n{\phi_n(x){\hat a}_n} $$ This gives immediately a null single-particle localization probability on the vacuum: $$ \rho_0(x) = \langle 0 | {\hat \psi}^\dagger(x) {\hat \psi}(x) |0 \rangle = \sum_{m,n}{\phi^*_m(x)\phi_n(x)\langle 0| {\hat a}^\dagger_m {\hat a}_n| 0\rangle } = 0 $$ In other words, the 2nd quantized vacuum is really … empty. In fact it is empty not only at the single-particle level, but at any k-particle level. Which means that the "probability of measuring a nonzero displacement in the particle's quantum field", as you asked, is really null. Or else, the theory does not provide the means to test the wavefunction of the vacuum state.
Note: one might think of a displaced or squeezed vacuum as a counterexample, but at a closer look nothing much changes because the situation simply transfers to unitarily transformed states and operators/observables.