It is not derivable from the Newton axioms as they are typically stated.
You need the extra assumption that your forces don't break rotation symmetry (as is pointed to by @rob's comment regarding the Noether theorem, which associates conserved quantities to symmetries).
Addendum
As an extended discussion about the derivation from Newtons axioms appeared under this post, let's just show what can be derived (and thereby see how one fails to derive conversation of total angular momentum without additional assumptions).
We have a system of $N$ particles at positions $\vec r_i(t)$ ($i = 1, \ldots, N$).
The equations of motion are (this is typically numberes as one of Newtons axioms):
$$m_i \ddot{\vec r} = \vec F_i(\vec r_1, \ldots, \vec r_n). $$
Now, we can look at the time-derivate of the total angular momentum:
\begin{align*}
\dot{\vec L} &= \partial_t \sum_{i=1}^N m_i \vec r_i \times \dot{\vec r_i} = \sum_{i=1}^N m_i \underbrace{\dot{\vec r_i} \times \dot{\vec r_i}}_{=0} + \sum_{i=1}^N\vec r_i \times (m_i\ddot{\vec r_i}) \\
&= \sum_{i=1}^N \vec r_i \times \vec F_i(\vec r_1, \ldots, \vec r_N)
\end{align*}
That is, we can derive a formula for the change of the angular momentum. The result is, however, not zero in general even if the forces obey $\text{actio} = \text{reactio}$.
As a counterexample, take a two particle system with:
\begin{align*}
\vec F_1(\vec r_1, \vec r_2) &= -\vec F_2(\vec r_1, \vec r_2) = \vec F(\vec r_2 - \vec r_1) \\
\vec F(\vec d) &= \alpha \vec e_x (\vec d \cdot \vec e_y)
\end{align*}
These forces clearly fulfil $\text{actio} = \text{reactio}$.
But the angular momentum is not constant in general, take the following initial conditions:
$$ \vec r_1(0) = \vec 0, \vec r_2(0) = l \vec e_y $$
Then (independently of the initial velocities), we have:
$$ \dot{\vec L}(0) = m_1 \alpha l \vec 0 \times \vec e_y - m_2 \alpha l \vec e_y \times \vec e_x = m_2 \alpha l \vec e_y $$
As the derivative does not vanish, the angular momentum is obviously not conserved.
If you take the strong version of the second axiom${}^1$ (as noted in a comment by @DavidHammen) then conservation of angular momentum does hold (the proof is left as an exercise for the reader ;) ). More general forces are permissible, as long as they are invariant under rotations).
${}^1$ That all forces act as action-reaction pairs, and are central, that is, point along the connecting line.
