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)
These quantities will be conserved by any system whose laws respect translational and rotational symmetry and are static in time. Respecting translational symmetry means that interactions don't depend on the positions of individual particles, just on differences of those positions. Respecting rotational invariance in turn means that interactions depend only on the magnitude of those differences. Being static means that the interactions don't depend explicitly on time or velocities. So this means that the two-point interaction must look like $V({\mathbf x}, {\mathbf y}) = f(|{\mathbf x - \mathbf y}|)$ (with $f$ suitably smooth). So yes, in particular your $r^{-n}$ potential will work.

Note that in general you could also consider more general interactions e.g. three- or more-point interactions. But to conserve the said quantities the interactions must still be singlet states of the respective groups.