Is the quantum number $L$ a combination of the azimuthal and magnetic quantum numbers $(\ell, m_\ell)$?
The values of $L$ come from the individual $\ell$ values, according to the Clebsch-Gordan series:
$$L = |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2.$$
where the dots indicate all &quot;in-between&quot; values with a step size of 1. The values of $S$ come from the individual $s$ values, in a similar fashion. Note that it is $s$, not $m_s$.
Finally, the values of $J$ come from $L$ and $S$:
$$J = |L - S|, \ldots, L + S.$$
The principal quantum number $n$ does not play any role in this, since it has no direct relation with angular momentum.

Usually it helps to see an example. Consider the case of the ground-state carbon atom; the $\mathrm{1s}$ and $\mathrm{2s}$ orbitals can be neglected since those are closed-shell, so we only need the two highest-energy electrons, i.e. a $\mathrm{p^2}$ configuration.
We therefore have two electrons, both having $\ell = 1$ (p-orbital) and $s = 1/2$ (because they are electrons, and all electrons are spin-$1/2$). The permissible values of $L$ are
\begin{align}
L &amp;= |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2 \\
  &amp;= | 1 - 1 |, \ldots, 1 + 1 \\
  &amp;= 0 , \ldots, 2 \\
  &amp;= 0, 1, 2
\end{align}
The permissible values of $S$ are
\begin{align}
S &amp;= |s_1 - s_2|, \ldots , s_1 + s_2 \\
  &amp;= | 1/2 - 1/2 |, \ldots, 1/2 + 1/2 \\
  &amp;= 0 , \ldots, 1 \\
  &amp;= 0, 1
\end{align}
The permissible values of $J$ depend on what $L$ and $S$ are, and must be evaluated on a case-by-case basis. So far, we have identified three possible values of $L$ and two possible values of $S$. Consider the case where $L = S = 0$: here,
\begin{align}
J &amp;= |L - S|, \ldots , L + S \\
  &amp;= |0 - 0|, \ldots, 0 + 0 \\
  &amp;= 0 , \ldots, 0 \\
  &amp;= 0
\end{align}
So when $L = S = 0$, the only permissible value of $J$ is also $0$. If you've seen term symbols before, the corresponding term symbol is denoted as $^1\!S_0$; if you haven't, then all you need to know is that this is a chemist's &quot;shorthand&quot; way of writing $(J, L, S) = (0, 0, 0)$, which is the combination we just found.
You will need to work through this for the other five combinations of $L$ and $S$. This will give you a full series of term symbols, except that there's a catch. The Pauli exclusion principle forbids some of those six combinations: to be precise, the combinations $(L, S) = (2, 1)$, $(1, 0)$, and $(0, 1)$ are not permissible, which leaves only three valid combinations, one of which I've already demonstrated. See Pauli-forbidden term symbols for atomic carbon for more details.

Addendum. As for the $m_\ell$ and $m_s$ quantum numbers: they do have a use, but in a rather more subtle way. I don't expect this very brief summary to be useful, but I'll try; feel free to ignore it at this stage of your learning if it doesn't make much sense.
Much like how (for one electron) $\ell = 1$ implies $m_\ell = -1, 0, 1$, for the entire atom each value of $L$ is associated with a range of $M_L$ values: for example, $L = 1$ implies $M_L = -1, 0, 1$. Therefore, the $L = 1$ term is actually not one single electronic state, but rather a collection of three different values of $M_L$, which is why we call it a &quot;term&quot; rather than a state. Likewise, the $L = 2$ term actually corresponds to five different values of $M_L$. How does this relate to $m_\ell$? Well, it turns out that $M_L$ is the sum of the individual electron $m_\ell$ values. A similar situation exists with $S$, $M_S$, and $m_s$. However, I think that's a topic for another question, if necessary.