I don't think your method is sound. Dimension-counting is a nice idea, but it doesn't really work. For example, nothing in your arguments prohibits the decomposition $$\mathcal{E}(n=2) = 3\mathcal{E}(n=2,l=0) \oplus \mathcal{E}(n=2,l=1),$$ where $\mathcal{E}(n=2)$ means the eigenspace of $H$ with $n=2$, and the informal notation $3\mathcal{E}(n=2,l=0)$ means the direct sum of three independent eigenspaces of $H$ and $L^2$ with $n=2$ and $l=0$ respectively. Moreover, this gets worse and worse at higher dimensions. You are arriving at your conclusions because you know what the results are to begin with.

This much is true, though:

I assumed that, if I have a state with a certain $l$ in $\mathcal{E}(n)$, then all the $2l+1$ states $\vert n,l,m\rangle$ are in $\mathcal{E}(n)$.

This is because $H$ commutes with all the components of $\mathbf L$, and you can get from any $|l,m\rangle$ to any other $|l,m'\rangle$ by applying suitable combinations of the components of $\mathbf L$. (This is most easily done, for those states, by applying the ladder operators $L_\pm = L_x \pm i L_y$.) More fundamentally, this is because the phrase "a subspace of definite $l$" really denotes an irreducible representation of the rotation group, and those are an everything-or-nothing kind of deal.

If you want an elegant method to resolve which $l$s contribute to the eigenspace at any given $n$, I would recommend the methods in this question or these notes.

I don't think your method is sound. Dimension-counting is a nice idea, but it doesn't really work. For example, nothing in your arguments prohibits the decomposition $$\mathcal{E}(n=2) = 3\mathcal{E}(n=2,l=0) \oplus \mathcal{E}(n=2,l=1),$$ where $\mathcal{E}(n=2)$ means the eigenspace of $H$ with $n=2$, and the informal notation $3\mathcal{E}(n=2,l=0)$ means the direct sum of three independent eigenspaces of $H$ and $L^2$ with $n=2$ and $l=0$ respectively. Moreover, this gets worse and worse at higher dimensions. You are arriving at your conclusions because you know what the results are to begin with.

This much is true, though:

I assumed that, if I have a state with a certain $l$ in $\mathcal{E}(n)$, then all the $2l+1$ states $\vert n,l,m\rangle$ are in $\mathcal{E}(n)$.

This is because $H$ commutes with all the components of $\mathbf L$, and you can get from any $|l,m\rangle$ to any other $|l,m'\rangle$ by applying suitable combinations of the components of $\mathbf L$. (This is most easily done, for those states, by applying the ladder operators $L_\pm = L_x \pm i L_y$.) More fundamentally, this is because the phrase "a subspace of definite $l$" really denotes an irreducible representation of the rotation group, and those are an everything-or-nothing kind of deal.

If you want an elegant method to resolve which $l$s contribute to the eigenspace at any given $n$, I would recommend the methods in this question or these notes.