wave function has $n_r$ nodes and the probability density $|\Psi|^2$ depends on $|\ell|$ and *not the sign* of $\ell$.

The corresponding eigenenergies are given by: $$E_{n_r,\ell} = \frac{\hbar\omega_c}{2}(2n_r+|\ell|+1) + \frac{\hbar \omega_c\ell}{2} =\hbar\omega_c\left(n_r+\frac{|\ell|+\ell}{2}+\frac{1}{2}\right),$$ so for all $\ell \leq 0 \ $, $E_{n_r,\ell} = \hbar\omega_c\left(n_r+\frac{1}{2}\right)$, and the energy spectrum is infinitely degenerate.

Sometimes a new quantum number $n \equiv n_r+{(|l|+l)}/{2}$ is defined, and then the energies are simply $$E_{n}=\hbar\omega_c\left(n+\frac{1}{2}\right).
    \label{energy_landau_level}$$ Clearly we see that further degeneracies are possible through the use of $\ell > 0$ for $\ell \leq n$. This is shown in the table below, where all energies have an infinite degeneracy due to the negative $\ell$ states that carry on indefinitely to the right for each energy. Additionally, however, as $n$ increases, positive $\ell$ states are also degenerate in growing number as indicated. $$\begin{aligned}
{\rm Energy} (\hbar \omega_c/2) \phantom{aaaaaaaa} n \phantom{aaaaaaaaaaaaaaaaaaaaa} (n_r,\ell) \ \ \phantom{aaaaaaaaaaaaaa} \cr
\vdots \phantom{aaaaaaaaaaaaaaaa} \vdots \phantom{aaaaaaaaaaaaa} \phantom{(0,2) \ \ (1,1) \ \ } \vdots \phantom{\ \ (2,-1) \ \ (2,-2)} \dots \cr
5        \phantom{aaaaaaaaaaaaaaaa} 2         \phantom{aaaaaaaaaa} (0,2) \ \ (1,1) \ \  (2,0) \ \ (2,-1) \ \ (2,-2) \dots \cr
3 \phantom{aaaaaaaaaaaaaaaa} 1        \phantom{aaaaaaaaaa} \phantom{(0,2)} \ \ (0,1) \ \  (1,0) \ \ (1,-1) \ 