I see two ways in which the thermal de Broglie wavelength is defined. In both cases we can get it from the probability distribution and partition function of an ideal gas. We will consider a 3D ideal gas with non-relativistic dispersion.
First Way
Consider the partion function of an ideal gas:
\begin{align}
Z &= \int_{p_x=-\infty}^{+\infty}\int_{p_y=-\infty}^{+\infty}\int_{p_z=-\infty}^{+\infty} e^{-\frac{1}{2mkT}(p_x^2+p_y^2+p_z^2)}dp_xdp_ydp_z\\
&= \left(2\pi mkT \right)^{\frac{3}{2}}
\end{align}
Note that this has dimensions of momentum cubed. Noticing this, we can define the characteristic thermal momentum
\begin{align}
p_T = \sqrt{2\pi mkT}
\end{align}
We can consider the de Broglie wavelength of a particle with this characteristic momentum to get the first definition of the thermal de Broglie wavelength:
$$\lambda_T = \frac{h}{p_T} = \frac{h}{\sqrt{2\pi mkT}}$$
Second Way
Consider the average energy of an ideal 3D gas. This can be found from the equipartion theorem to be
\begin{align}
\langle E \rangle = \frac{3}{2}kT
\end{align}
A relation for the de Broglie wavelength is
$$
\lambda = \frac{h}{\sqrt{2mE}}
$$
Considering the de Broglie wavelength of a particle with energy equal to the average thermal energy of a 3D ideal gas we get the second definition of the thermal de Broglie wavelength:
$$
\lambda_T = \frac{h}{\sqrt{3mkT}}
$$
A Note on a Possible Third Way
A third way which would make sense would be to calculate the average de Broglie wavelength of all of the particles in an ideal gas:
$$
\langle \lambda \rangle = \iiint \frac{h}{p} e^{-p^2/2mkT} dp^3 = \frac{2h}{\sqrt{2\pi m kT}}
$$
We see that this is within a factor of 2 of the first definition.
Summary
The three approaches differ by factors of order unity so they all refer to similar length scales. In the end the thermal de Broglie wavelength is largely a notational convenience so we don't need to carry around factors of $\frac{h^2}{mkT}$ all over the place so we shouldn't worry too much about the pre-factor. But it is nice to know where the different conventions come from. Though it is largely a notational convenience it does clearly have physical significance since it is related to $\langle \lambda \rangle$.
I have never really seen the third way presented. I have seen the first way presented by far the most often. I think this is because the partition function appears all over the place so very commonly the specific factor $2\pi mkT$ arises so it is nice to give this quantity a name. The second approach may be presented more often in introductory approaches to statistical mechanics. This convention is most problematic because it very clearly relies on the specific problem of a 3 dimensional ideal gas.
Best Answer
A more general principle in the quantum framework is the Heisenberg uncertainty relation
$$\large{\color{red}{\Delta \mathrm{x}\Delta \mathrm{p}\ge \frac{\hbar}{2} \\ \Delta \mathrm{E}\Delta \mathrm{t}\ge \frac{\hbar}{2} }}$$
It tells us that when the momentum is zero the position is indeterminate, actually it could go from zero to infinity to obey the principle.
This is consistent with the wavelength being infinite in the de Broglie relation, and it means that the localization of the particle is indeterminate.
What do these generalized relations mean? The Heisenberg uncertainty principle is in one to one correspondence with the quantum mechanical commutator of the two observables. It tells us that if the wavefunction for the problem under consideration is acted upon by the momentum operator and a momentum eigenvalue is obtained the position is indeterminate. It is probable to find the particle anywhere within the boundary conditions, according to the probability distribution given by the square of the wave function.
As we know that quantum mechanics applies to the dimensions of molecules and atoms, and we do observe them in fixed within a width locations in a crystal, and in scattering experiments with fixed momentum beams the particle again are localized within nanometer widths, it is obvious that the simple sinusoidal waves are not adequate to model matter. One uses wave packets to do that, which give localization and can model a particle .
In other words a single frequency plane wave, which is what the de Broglie relation implies is not a good model for physical matter particles.