The strong force has an approximate symmetry called isospin, which is a rotation in an abstract space spanned by the neutron and proton. This symmetry is broken by the Coulomb force, which only acts on protons, and by the slight mass difference between up quarks and down quarks, which makes the neutron slightly heavier than the proton.
Consider possible bound states of neutrons and protons. These are fermions, so the wave function has to be anti-symmetric. Neutrons and protons are heavy, so we can analyze the situation using non-relativistic quantum mechanics. If a bound state exists, the ground state is in an s-wave (symmetric) because the centrifugal barrier is purely repulsive. This means that the spin-isospin wave function has to be anti-symmetric. This leaves two possibilities:
$$
(I=0,S=1) \;\;or\;\; (I=1,S=0)
$$
where $I$ and $S$ are the total isopsin and spin of the state. Note that $I,S=0$ is anti-symmetric in isospin/spin, and $I,S=1$ is symmetric. The $I=0$ state corresponds to
$$
|I=0\rangle = \frac{1}{\sqrt{2}}\left(|np\rangle -|pn\rangle\right)
$$
and $I=1$ has three isospin components
$$
|I=1,I_3=+1,0,-1\rangle = \left\{|pp\rangle ,\frac{1}{\sqrt{2}}\left(|np\rangle +|pn\rangle\right),|nn\rangle\right\}
$$
The potential in the $I=0,1$ channels is determined by complicated interactions between quarks and gluons.
Empirically (or from numerical calculations in lattice QCD) we know that both the $I=0$ potential and the $I=1$ potential are attractive, but the $I=0$ interaction is slightly more attractive.
Note that in non-relativistic QM (in three dimensions), an attractive interaction is not sufficient to form a bound state. The potential has to have a minimum strength. Nuclear physics is fine-tuned, the empirical NN potentials are close to the threshold where bound states appear.
There is an $I=0$ bound state, called the deuteron (binding energy 2.2 MeV), but the di-neutron (and its isospin partners) fall just short of being bound. Due to the Coulomb force, there is a little extra repulsion in $pp$, but the $nn$ channel is almost bound.
This can be quantified using the scattering length. The $nn$ scattering length is about 20 fm, and the corresponding energy scale is
$$
B=\frac{1}{2ma^2}\simeq 50 \, keV
$$
so the di-neutron comes within 50 keV of being bound. This also means that if one could change the quark masses slightly, there would presumably be a bound di-neutron (this can be checked in lattice QCD).
It is possible in principle to have a three neutron bound state even if there is no two neutron bound state (states like that are called Borromean), but this is disfavored by the Pauli principle (not all three neutrons can be in an s-wave). The same applies to 4n bound states (there is a recent claim of a 4n resonance http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.052501).
In practice $^3H=pnn$ and $^3He=ppn$ exist, but not $3n$.
Note that neutron stars, clusters of $10^{57}$ neutrons, do exist. Of course, these objects are gravitationally bound.
Best Answer
Suppose you consider two satellites orbiting a planet with their orbits aligned at right angles:
This arrangement of the three objects produces a three dimensional object, even with just three points, because the plane formed by the three objects continuously changes its orientation in 3D space.
The above shows that even for a purely classical system three objects can define a 3D structure. For completeness we should point out that quantum objects, like quarks in a nucleus or electrons in an atom, are not points and do not have a position in the sense that a classical object has a position. Quantum objects are delocalised in space, a bit like a fuzzy cloud, so they are inherently three dimensional. A hydrogen atom does not consist of a pointlike electron orbiting in a plane around the proton. The electron in a hydrogen atom is a fuzzy 3D blob centred on the proton.
And to make things even more complicated, hadrons like protons do not consist of three quarks. For the quarks (and gluons) inside a hadron the concept of discrete particles is somewhat vague and a hadron is really a roiling sea of an indeterminate number of particles. When we say a hadron contains three quarks we mean only that on average there are three quarks present.