There are already two good answers but I have the feeling that something is still missing. In fact, consider a system of more than 2 particles. The force on the particle $1$ is

$$\textbf{F}_{\text{1 due to 2..N}}=\textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) \, .$$

Generally, this is a non-linear function of the position $\textbf{r}_1$ given the positions $\textbf{r}_2,..,\textbf{r}_N$. Apart from the positions, you may also want to consider the velocities or other degrees of freedom (spin maybe) of the particles if you want.

Therefore, $\textbf{F}_{\text{1 due to \, 2..N}}$ "looks like" an $N$-body force, but typically $\textbf f$ can be decomposed as

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) $$

If you can do this, then $\textbf f$ is not a "genuine" $N$-body force, but rather it is a sum of 2-body forces.

You are forced to consider 3-body forces if you discover that the above decomposition is impossible, but that the physical system is well described by

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) + \sum_{i<j=3...N} \textbf{h}(\textbf{r}_1|\textbf{r}_i,\textbf{r}_j) $$

In this case the $ \textbf{h}$ are your "genuine three body forces", i.e. it is impossible to split $ \textbf{h}$ as $ \textbf{h}(1|ab)=\textbf{m}(1|a)+\textbf{m}(1|b)$, where $\textbf{m}$ is a two-body interaction (possibly different from $\textbf{q}$).

Now, the question why terms like $\textbf{h}(1|ab)$ are theoretically possible but "not observed" (if someone knows examples that do not involve QCD or quantum stuff please comment) observed in the "macroscopic" world is interesting and I do not have a complete answer for that. However, my feeling is that it has something to do with the kind of interactions that mediate the "action at a distance". For example, in celestial mechanics the gravitational potential is a linear superposition of the single-particle potentials: this ensures the two-body nature of the gravitational attraction. The same for Coulomb forces. Since General Relativity is non-linear, I suspect that there could be some post-Newtonian correction to the motion of a system of "planets" that gives rise to tree-body forces (but maybe those corrections are so small that are never considered).

There are already two good answers but I have the feeling that something is still missing. In fact, consider a system of more than 2 particles. Formally, the force on the particle $1$ is

$$\textbf{F}_{\text{1 due to 2..N}}=\textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) \, .$$

In general, this is a non-linear function of the position $\textbf{r}_1$ given the positions $\textbf{r}_2,..,\textbf{r}_N$. Apart from the positions, you may also want to consider the velocities or other degrees of freedom (spin maybe) of the particles, but this is not fundamental to discuss the point.

Therefore, $\textbf{F}_{\text{1 due to 2..N}}$ looks like a $N$-body force, but typically $\textbf f$ can be decomposed as

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) $$

If you can do this, then $\textbf f$ is not a "genuine" $N$-body force, but rather it is just the total force due to the sum of 2-body interactions.

You are forced to consider 3-body forces if you discover that the above decomposition is impossible (or if you discover that you cannot reproduce the observed dynamics of your system bu considering only two-body interactions). If this is the case, your physical system may be well described by considering a more general decomposition:

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) + \sum_{i<j=3...N} \textbf{h}(\textbf{r}_1|\textbf{r}_i,\textbf{r}_j) $$

In this case the $ \textbf{h}$ are your "genuine three body forces", i.e. it is impossible to split $ \textbf{h}$ as $ \textbf{h}(1|ab)=\textbf{m}(1|a)+\textbf{m}(1|b)$, where $\textbf{m}$ is a two-body interaction (possibly different from $\textbf{q}$).

Now, the question why terms like $\textbf{h}(1|ab)$ are theoretically possible but "not observed" (if someone knows examples that do not involve QCD or quantum stuff please comment) observed in the "macroscopic" world is interesting and I do not have a complete answer for that. However, my feeling is that it has something to do with the kind of interactions that mediate the "action at a distance". For example, in celestial mechanics the gravitational potential is a linear superposition of the single-particle potentials: this ensures the two-body nature of the gravitational attraction. The same for Coulomb forces. Since General Relativity is non-linear, I suspect that there could be some post-Newtonian correction to the motion of a system of "planets" that gives rise to tree-body forces (but maybe those corrections are so small that are never considered).

There are already two good answers but I have the feeling that something is still missing. In fact, consider a system of more than 2 particles. The force on the particle $1$ is

$$\textbf{F}_{1 \text{due to 2..N}}=\textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) \, ,$$

namely it is (in general) a non-linear function of the position $\textbf{r}_1$, given the positions $\textbf{r}_2,..,\textbf{r}_N$. Add the velocities or other degrees of freedom (spin maybe) of the particles if you want. This "looks like" an $N$-body force, but typically $\textbf f$ can be decomposed as

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) $$

Is you can do this, then $\textbf f$ is not a "genuine" $N$-body force, but rather it is a sum of 2-body forces. You are forced to consider 3-body forces if you discover that the above decomposition is impossible and that you are forced to consider

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) + \sum_{i<j=3...N} \textbf{h}(\textbf{r}_1|\textbf{r}_i,\textbf{r}_j) $$

In this case the $ \textbf{h}$ are your "genuine three body forces", i.e. it is impossible to split $ \textbf{h}$ as $ \textbf{h}(1|ab)=\textbf{m}(1|a)+\textbf{m}(1|b)$, where $\textbf{m}$ is a two-body interaction (possibly different from $\textbf{q}$). Now, the question why terms like $\textbf{h}(1|ab)$ are theoretically possible but "not observed" (if someone knows examples that do not involve QCD or quantum stuff please comment) observed in the "macroscopic" world is interesting and I do not have a complete answer for that. However, my feeling is that it has a lot to do with the kind of interactions that mediate the "action at a distance". For example, in celestial mechanics the gravitational potential is a linear superposition of the single-particle potentials: this ensures the two-body nature of the gravitational attraction. The same for Coulomb forces.

There are already two good answers but I have the feeling that something is still missing. In fact, consider a system of more than 2 particles. The force on the particle $1$ is

$$\textbf{F}_{\text{1 due to 2..N}}=\textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) \, .$$

Generally, this is a non-linear function of the position $\textbf{r}_1$ given the positions $\textbf{r}_2,..,\textbf{r}_N$. Apart from the positions, you may also want to consider the velocities or other degrees of freedom (spin maybe) of the particles if you want.

Therefore, $\textbf{F}_{\text{1 due to \, 2..N}}$ "looks like" an $N$-body force, but typically $\textbf f$ can be decomposed as

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) $$

If you can do this, then $\textbf f$ is not a "genuine" $N$-body force, but rather it is a sum of 2-body forces. 

You are forced to consider 3-body forces if you discover that the above decomposition is impossible, but that the physical system is well described by

$$ \textbf{f}(\textbf{r}_1|\textbf{r}_2,..,\textbf{r}_N) = \sum_{j=2...N} \textbf{q}(\textbf{r}_1|\textbf{r}_j) + \sum_{i<j=3...N} \textbf{h}(\textbf{r}_1|\textbf{r}_i,\textbf{r}_j) $$

In this case the $ \textbf{h}$ are your "genuine three body forces", i.e. it is impossible to split $ \textbf{h}$ as $ \textbf{h}(1|ab)=\textbf{m}(1|a)+\textbf{m}(1|b)$, where $\textbf{m}$ is a two-body interaction (possibly different from $\textbf{q}$). 

Now, the question why terms like $\textbf{h}(1|ab)$ are theoretically possible but "not observed" (if someone knows examples that do not involve QCD or quantum stuff please comment) observed in the "macroscopic" world is interesting and I do not have a complete answer for that. However, my feeling is that it has something to do with the kind of interactions that mediate the "action at a distance". For example, in celestial mechanics the gravitational potential is a linear superposition of the single-particle potentials: this ensures the two-body nature of the gravitational attraction. The same for Coulomb forces. Since General Relativity is non-linear, I suspect that there could be some post-Newtonian correction to the motion of a system of "planets" that gives rise to tree-body forces (but maybe those corrections are so small that are never considered).