Question: Given a collection of point-particles, interacting through an attractive force $\sim \frac{1}{r^2}$.

Knowing only $m\_1a=\sum\_i \frac{Gm\_1m\_i}{r^2}$ and initial conditions we can deduce the motion of the system.

Consequently we can observe that three quantities remains constant
A) center of mass of the system
B) total energy
C) angular momentum

How can we derive these 3 facts directly from $m\_1a=\frac{Gm\_1m\_2}{r^2}$ ?

Are these quantities conserved for any attractive force $\sim\frac{1}{r^n}$ ?

Given any monotonically decreasing force for $r$ in $(0,\infty)$, which are the conserved quantities?

Answer: 
Actually, this is the basic stuff every mechanics textbook should have. 

1. Center of mass is the most basic it needs just Newton laws:  

Second: $m\_i\ddot{\vec{r}\_i}= \vec{F}\_i$  

And third: $\sum\_i\vec{F}\_i = 0$  

Summing over i one obtains:  

$\frac{d^2}{dt^2}\left(\sum\_i m\_ir\_i\right) = \sum\_i\vec{F}\_i = 0$
2. For the total energy you need those forces to be potential and independent on time:  

$F\_i = -\frac{d}{d\vec{r}\_i}U(\vec{r}\_1,\vec{r}\_2,...)$  

Then just take the total energy :
$E = \sum\_i \frac{m|\vec{r}\_i|^2}{2}+U$  

And differentiate with respect to time:  

$\frac{dE}{dt} = \sum\_i m\_i(\ddot{\vec{r}}\_i\dot{\vec{r}}\_i)+\sum\_{i} \frac{dU}{d\vec{ r}\_i}\dot{\vec{r}}\_i = 0$
3. Finally, for the angular momentum $U$ must be rotational invariant:  

$U(R\vec{r}\_1,R\vec{r}\_2,R\vec{r}\_3,...) = U(\vec{r}\_1,\vec{r}\_2,\vec{r}\_3,...)$  

where R is the rotation matrix. Consider now very small (infinitesimal) rotation:  

$R\vec{r}\_i = \vec{r}\_i+\left[\delta\vec\phi\times\dot{\vec{r}}\_i\right]$  

Substituting and expanding, one can get:  

$U+\delta\vec\phi\sum\_i\left[\frac{dU}{d\vec{r}\_i}\times\vec{r}\_i\right]=U$  

Which works for every angle $\delta\phi$, so the following must hold
$\sum\_i\left[\frac{dU}{d\vec{r}\_i}\times\vec{r}\_i\right]=0$ 

Finally take the angular momentum:  

$\vec{M} = \sum\_i m\_i \left[\vec{r}\_i\times\dot{\vec{r}}\_i\right]$  

And differentiate with respect to time:  

$\frac{d\vec{M}}{dt} = \sum\_i \left[\vec{r}\_i\times m\_i\ddot{\vec{r}}\_i\right]+
\sum\_i m\_i \left[\dot{\vec{r}}\_i\times \dot{\vec{r}}\_i\right] = \sum\_i\left[\frac{dU}{d\vec{r}\_i}\times\vec{r}\_i\right]=0$

Phew... You can see that I never used the "pairwise" interactions. And I only needed the basic assumptions about the potential energy. 

Concerning "extra" things that you cans say for your systems -- check [the Virial theorem](http://en.wikipedia.org/wiki/Virial_theorem)
The total momentum is also conserved.

Noether's theorem relates conserved quantities to continuous symmetries of the system.

For the inverse square law, there's also a less known symmetry
$x \to e^\lambda x$, $t \to e^{2\lambda} t$.

But we shouldn't expect any more because of chaos; this isn't an integrable system.