What exactly causes equilibrium to shift when pressure is decreased? Take the reaction:

$$\ce{N2(g) + 3 H2(g) &lt;=&gt; 2 NH3(g)}$$

If pressure is decreased, equilibrium shifts to the left because it has more moles. I understand the basics of why this is, i.e. due to Le Chatelier's principle, but is there an actual explanation as to why the system does this (e.g. using collision theory)? What's actually happening to cause the shift? 

My chemistry teacher says she doesn't know of an explanation.
This happens because the equilibrium constant $K_p$ is independent of the pressure, although it does depend on temperature, which we will assume to be constant. The equilibrium constant can be written as 

$$\displaystyle K_p =\frac{p_A^2}{p_Np_H^3}$$

where for simplicity I have abbreviated the chemical names. The partial pressure $p$ can be written in terms of the degree of dissociation and total pressure $P_T$, i.e. $p=aP_T$ where $a$ is a mole fraction term in the degree of dissociation. As partial pressure is proportional to the degree of dissociation and total pressure, for example, $p_A =a_A^2  P_T^2$ then the equilibrium constant 

$$K_p=\frac{a_A^2}{a_Na_H^3} \frac{1}{P_T^2}$$

where the terms in $a$ we have not yet evaluated and contain the degree of dissociation $\alpha$. However, you can see that as $K_p$ is constant if $P_T$ decreases then $1/P_T^2$ increases and so  $\displaystyle \frac{a_A^2}{a_Na_H^3} $ must decrease. This means that the degree of dissociation decreases and so the equilibrium 'moves to the left'.

Working out the partial pressure for this reaction is a bit messy. 

Let $\alpha $ be the degree of dissociation then $(1-\alpha)$ of nitrogen reacts , $3(1-\alpha)$ hydrogen and $2\alpha$ ammonia is produced. The total of these is $6-2\alpha$ and so the mole fraction of ammonia is, for example, $a_A=\displaystyle \frac{2\alpha}{6-2\alpha}$ and the partial pressure $p_A=\displaystyle \frac{2\alpha}{6-2\alpha}P_T$. You can make a similar expression for the other species and substitute to get the mole fraction term $\displaystyle \frac{a_A^2}{a_Na_H^3}=\frac{16\alpha^2(\alpha-3)^2}{27(\alpha-1)^4}$. This expression increases as $\alpha$ increases. As you see the expression is messy but the main point is that as the pressure decreases the mole fraction term must decrease which it does by reducing $\alpha$.