Conserved quantities in generalized n-body problem 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?
 A: Actually, this is the basic stuff every mechanics textbook should have.  


*

*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:
$\frac{d^2}{dt^2}\left(\sum_i m_ir_i\right) = \sum_i\vec{F}_i = 0$

*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:
$\frac{dE}{dt} = \sum_i m_i(\ddot{\vec{r}}_i\dot{\vec{r}}_i)+\sum_{i} \frac{dU}{d\vec{ r}_i}\dot{\vec{r}}_i = 0$    

*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
A: 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.
A: 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.
A: There are many more conserved quantities in a potential N-body problem but not all of them are additive on particles. This fact is not related to the form of potentials at all.
A: All codes of the N-body simulation make use of instantaneous propagation of the speed of gravity.
The other answers follow the same criterion that is good enough for local simulations.
As a consequence of the finite speed of propagation of gravity and the motion of the masses there is a dissipative component.
As the bodies are never static, the masses continuously lose energy to space because there is no known mechanism that reverses the direction of flow.
(An equivalent effect applies to the radiation of charged particles by virtue of its continued accelerated motion.)
from Jorgen Kalckar and Ole Ulfbeck experiment (a) (energy of two masses and a spring):
...

The difference between these two
  energies is lost by each mass: it is
  taken away by space-time, in other
  words, it is radiated away as
  gravitational radiation
gravity can behave like a wave:
  gravity can radiate
  ...
  All this follows from the expression of universal gravity when applied to moving observers, with the requirement that neither observers
  nor energy can move faster than c.

(a) motionmountain  free online book, ch 18, pag 526/527: Motion in General Relativity, Gravitational Waves.
