Conservation of momentum follows immediately from Newton's 2nd and 3rd law. For simplicity, I will assume a 2-particle system but the logic can be easily generalised to many particles. For the purpose of the derivation I will also assume that the only force acting on particles is the intermolecular forces between them.
Given two particles (1 and 2) in 3-d, the equations of motion (via Newton's 2nd Law) is given by
\begin{align}
\frac{d\vec{p_1}}{dt} = \vec{F_1},\hspace{0.5cm} \frac{d\vec{p_2}}{dt} = \vec{F_2}
\end{align}
where $\vec{F}_1$ is the force acted on the particle 1 by particle 2 and vice versa for $\vec{F}_{2}$. By Newton's Third Law, we know $\vec{F}_1 = - \vec{F_2} $ so we have
\begin{align}
\frac{d\vec{p_1}}{dt} + \frac{d\vec{p_2}}{dt} = \vec{F_1} +\vec{F_2} = \vec{F_1} - \vec{F_1} = 0
\end{align}
Collecting things up, we get
\begin{align}
\frac{d}{dt}(\vec{p_1} + \vec{p_2}) = 0
\end{align} meaning the total momentum of the system doesn't change.