For ideal classical gasses, in terms of the energy levels why do we ignore whether the particles are fermions or bosons?

Because both the Fermi-Dirac distribution and the Bose-Einstein distribution are well approximated by the Maxwell-Boltzmann distribution in the limit of low density. Basically, the assumption behind the ideal gas is that the gas density is low enough that collisions are not a significant factor in describing the dynamics of the gas, allowing us to go from the thermodynamic distribution of 1 particle to the gas. When density starts becoming high, the first correction is usually described by the Van der Waals equation. As the temperature drops or density goes up further, you have to start worrying about the Boson/Fermion distinction.

More precisely, it's not about $kT$ compared to some energy level spacing, it's compared to the chemical potential. In detail, it's having $kT$ high enough that \begin{align} \mathrm{BE}(E) & = \frac{1}{\operatorname{e}^{(E-\mu)/kT}-1} \mathrm{\ and} \\ \mathrm{FD}(E) & = \frac{1}{\operatorname{e}^{(E-\mu)/kT}+1} \end{align} are adequately approximated by \begin{align} \mathrm{MB}(E) & = \operatorname{e}^{-E/kT}. \end{align}

Note that the approximation is only "good" at the low end ($E<kT$) when $\mu<0$.

For Fermions, the chemical potential is the Fermi energy or greater, which is controlled by the particle density. I'm having trouble finding a reference for how to find the chemical potential for the Bose-Einstein distribution. ResearchGate hosts a plot of the chemical potential of Heliums 3 and 4 at low temperature ($\mu/k$ is around $-2$ and $-7$ Kelvin for them, respectively).

Another way of seeing it is that the distance between atoms is large compared to their de-Broglie wavelength. Then it does bot matter that one should use statistics of indistinguishable particles - it would in principle still be possible to follow a particle most of the time.

There are already good answers; I'll just add another way to see this. Let $n$ be the number of particles in a particular quantum state. Using Maxwell-Boltzmann statistics, you can calculate a probability distribution $p_{\text{MB}}(n)$.

Fermions modify this distribution by forbidding more than one particle in the same state, $$p_{\text{Fermi}}(2) = p _{\text{Fermi}}(3) = \ldots = 0.$$ Bosons modify this distribution by preferring to 'clump up', i.e. one tends to see groups of bosons in the same state, $$p_{\text{Bose}}(2) \gtrsim p_{\text{MB}}(2), \quad p_{\text{Bose}}(3) \gtrsim p_{\text{MB}}(3), \ldots$$ In all cases, the average $\langle n \rangle$ is the same since there are the same number of total particles.

The limit where all of these distributions are the same is the low-density limit $\langle n \rangle \ll 1$. In this case, the overwhelming majority of the probability is concentrated in $p(0)$ with a tiny bit in $p(1)$. The modifications that the Fermi and Bose distributions make to $p(2)$ and higher are negligible.

Since there's one quantum state for every Planck's constant of phase space area, $\langle n \rangle \ll 1$ is equivalent to $$(\text{typical momentum})(\text{typical distance between neighboring particles}) \gg h.$$ As already mentioned, this is equivalent to saying that the particles are separated by a distance much greater than their de Broglie wavelength.