The word temperature usually refers to the average velocity of massive particles, correct?
And the Cosmic Microwave Background (CMB) has a 'temperature' based on the temperature of a 'black body' that would emit photons of energies corresponding to those seen in the CMB, correct?
But, how can a neutrino or neutrinos have a temperature? What does it correspond to?
The temperature of a gas is a parameter that reflects the distribution of energy/momentum of the particles. It is not a characteristic of any individual particle.
Before the cosmic neutrino background was formed (when the early universe was $>10^{11}$ K) neutrinos and anti-neutrinos were produced and destroyed in thermal equilibrium with the rest of the radiation and baryonic matter. That is, the neutrinos had a distribution of energies and momenta that was determined by the temperature of the universe at that time. NB: This is not a blackbody distribution, it is the Fermi-Dirac distribution because neutrinos are spin 1/2 particles with mass.
As the universe expanded and cooled, the density fell, and at about 1 second after the big bang, the interaction timescale for the neutrinos became longer than the expansion timescale of the universe. The neutrinos "decoupled" from the other matter and radiation, but their distribution of momentum was preserved, with a characteristic temperature of a few $10^{10}$ K.
Since then, the universe has expanded by a factor of $\sim 10^{10}$ and the momentum of the individual neutrinos with respect to the comoving frame has decreased by a similar amount. (Even though the neutrinos have a small mass, you can think of the process as the expansion stretching their de Broglie wavelengths). Thus the neutrinos still have a momentum distribution, but it is now the equivalent of a much colder gas - about 2K.
A single particle is not assigned a temperature (which is exhibited by a very large ensemble of particles), it is described by its kinetic energy. So a single neutrino can be described as having a certain amount of kinetic energy- but it in and of itself has no "temperature".
Now if we imagine instead a huge burst of neutrinos released during a supernova collapse inside a supermassive star, that burst will contain a range of kinetic energies which start out being all characteristic of the process which created them and then when the neutrinos interact with themselves and with the matter and radiation surrounding them in the core of that star, that distribution will get averaged into a blackbody distribution with a peak to which a temperature can be ascribed.
What is the cosmic microwave background radiation "temperature"?
It is the value of temperature found when fitting a black body radiation curve to its spectrum of photons.
The perfect fit leads to the present cosmological model's hypothesis that it is the result of the loss of energy of the original spectrum from the decoupling of radiation from matter in the Big Bang expansion.
So the temperature of the CMB reflects the kinetic energies of the particles in the plasma interacting at the time the radiation decoupled.
In an analogous analysis, the measured spectrum of cosmic neutrinos will have information about the temperature they were produced in the history of the universe. If interested further see this. .
So the temperature of a radiation refers to the black body fit temperature of the particles that produced it.
Loosely speaking, any system where the degrees of freedom can exchange energy with each other and is in equilibrium has a temperature.
A "system" here is a collection of particles, fields, etc. that you can draw a conceptual box around. "Equilibrium" means the populations of the states that can exchange energy with each other are not varying in time. When that's true, the relative probability of two states is the famous Boltzmann Factor, $\exp \left( -\frac{\Delta E}{k_\text{B} T} \right)$.
A consequence is you can sum $\exp \left( \frac{E}{k_\text{B} T} \right)$ over all possible configurations of the system (positions, velocities, numbers of neutrinos of all energies, etc.) to get the partition function $Z$.
Then you can prove that energy $U$, entropy $S$, heat capacity, and so on are all mathematically computable from $Z$. I'd argue in my (somewhat) humble opinion that $Z$ is the cornerstone of thermodynamics. Most everything else is
The confusion in themodynamics is the clash between the last two. Our modern understanding derives macroscopic properties from microscopic ones. But historically we saw the macroscopic ones first and invented empirical relationships among them. Science tends to be taught in historical order for some reason so we learn special cases before the general theories.
The word temperature usually refers to the average velocity of massive particles, correct?
So with that long tangent out of the way: you're right that it's usually understood that way. But we now also see that velocity of particles is just one property among many that contribute to the total number of states and their energies. It's just part of the overall story. Sometimes a negligible part. So in general the statement "temperature refers to the average velocity of massive particles" is false. Sometimes it's true. Other times, like in blackbody radiation, velocity is irrelevant.
Do individual particles have a temperature? Rarely. Temperature is a property of many individual states being in equilibrium. Complicated particles with many degrees of freedom could have their own temperature if those states exchange energy among themselves. But most microscopic particles are simple, with few degrees of freedom, and those degrees of freedom don't exchange energy among themselves. Usually it takes a lot of particles to build up a statistical ensemble. Then you have enough states to have a meaningful average energy of the states and an equilibrium.
A collection of neutrinos can therefore therefore have a temperature when they're at thermal equilibrium. They're like any other system, it's just that the specific degrees of freedom and interactions that are different. But the overall phenomenon of microstates and exchanging energy and an equilibrium with the Boltzmann factor is the same!
