A neutrino has rest-mass and travels at (near) $c$, why isn't its mass/ energy (nearly) infinite?
Because it has too low rest mass or still too low speed. Neutrinos are very light particles: Their rest energy is comparable to energy of a hydrogen bond (weaker than typical chemical bound). So you can understand, that full energy must not be infinite.
However, this is not a practical point of view. From the practical point of view, neutrinos have mass-energy as they have (because they got it when they were born (or sometimes in an interaction event afterwards)) and because of this they have speed (that for all currently observable neutrinos is only slightly less than the speed of light).
The mass-energy can be anything bigger than the rest mass of neutrino. Neutrino detector using a gallium → germanium transformation has detection threshold of 0.233 MeV and this is the lowest value I have found in the Wikipedia article about neutrino detection. Solar neutrinos have energies up to 18 MeV. Famous, but now known to be mistake, faster-than-light neutrino anomaly has something to do with 28 GeV neutrinos. 6.5 TeV per beam (planed for LHC since early in 2015) can become almost wholly energy of neutrino. (Two protons could stop and emit neutrino and antineutrino, but probability is low, digression 1).
The equation for energy of particle with rest mass $m$ is
$E = \gamma m$
where
$\gamma = \frac{1}{\sqrt{1-v^2}}$ so $v = \sqrt{1-\frac{1}{\gamma^2}} \approx 1 - \frac{1}{2 \gamma^2}$.
In the cases about which I write above, the speed is respectively $(1-9.2 \cdot 10^{-14}) c$, $(1-1.5 \cdot 10^{-17}) c$, $(1-3.4 \cdot 10^{-24}) c$ and $(1-1.2 \cdot 10^{-28}) c$.
On the other hand, "relic background neutrinos are estimated to have (...) temperature 1.9 K ($1.7×10^{−4}$ eV) if they are massless, much colder if their mass exceeds 0.001 eV". So even assuming temperature too high for such "heavy" 0.1 eV neutrinos, they are non-relativistic and have average speed (from non-relativistic equations $E = m v^2 / 2$ and $E = \frac{3}{2} k T$) $v = \sqrt{3 k T / m} \approx 0.071 c$. That's still about 21000 km/s, but clearly less than the speed of light.
So if we plug in to find the total energy of the neutrino we find.
ev∼18keV∼0.03 $m_{electron} Which isn't that big.
Ia this answer correct?
J.J. must have forgotten about square. I have corrected it and $0.03 m_{electron}$ is now even better approximation, but this says something only about neutrinos with speed $0.999999997 c$.
Now, the energy of an electron is .5 Mega eV, what is then the correct maximum value of the energy of a neutrino?
It is true that "the mass of the neutrino is tiny, but its kinetic energy can be of the same scale as that of the electron" for neutrinos from beta decay and the energy of beta decay is comparable to the mass of electron, so this question makes some sense, but knowing just this, one can only say "of the order of .5 Mega eV". Solar neutrinos are from nuclear reactions and have energies up to 18 MeV - the same order more or less.
You probably wanted some calculations, so I will calculate maximum energy of the neutrinos from decay of tritium. In the first approximation, energy of neutrino is simply difference between rest energy of tritium atom and rest energy of helium-3 atom. (If neutrino takes the whole energy, we could really get neutral helium-3 atom, but probability is low. Digression 2.) Neglecting neutrino mass, its energy is equal to its momentum ($p$), which is equal (with opposite direction) to the helium-3 momentum.
$E_{{^3}\mathrm{He}}^2 = m_{{^3}\mathrm{He}}^2+p^2$
$m_\mathrm{T} = E_{{^3}\mathrm{He}} + p = \sqrt{m_{{^3}\mathrm{He}}^2+p^2} + p$
$m_{{^3}\mathrm{He}}^2 + p^2 = m_\mathrm{T}^2 - 2 m_\mathrm{T} p + p^2$
$p = \frac{m_\mathrm{T}^2 - m_{{^3}\mathrm{He}}^2}{2 m_\mathrm{T}}$
$p = 0.0186 \mathrm{M}e\mathrm{V}$
using
$1 \mathrm{u} = 931.4812 Me\mathrm{V}$
$m_\mathrm{T} = 3.01604927 \mathrm{u}$
$m_{{^3}\mathrm{He}} = 3.01602931 \mathrm{u}$
In this case, we get low energy. From 0.0186 MeV to 18 MeV is something about 0.5 MeV, but writing this I have realised that the rule that nuclear reaction energy is comparable to the mass of electron is not very precise.
Digression 1: Probability is low, but I do not know how low - there are very many events and I do not have any idea, how often this is going to happen - maybe more than once a second, maybe less than once per trillion years. This should be quite easy to estimate if one knows appropriate rule, but I don't know it and I am afraid that looking for it would take too much time. This has something to do with parton energy distribution and probability of reaction. And what does "almost wholly" mean? Something like 99%, in worst case 99.9%, is enough to get $(1-1.2 \cdot 10^{-28}) c$ with 1.2, not 1.3. Anyway, taking one half of this energy should be much more probable and give comparable neutrino speed.
Digression 2: Again I do not know how improbable this is. This should be quite easy to estimate if one knows appropriate rule, but I don't know it and I am afraid that looking for it would take too much time. This has something to do with distribution of speed in decays, but also with something like resonances.
General notes:
I do not know more than Wikipedia says about the 0.320 ± 0.081 eV estimation by Planck collaboration, but here I'm assuming that it is correct. Than, oscillation data show that the mass difference is small, and masses of all three neutrinos are about 0.1 eV, but this is rough approximation from error range itself, so I probably underestimate other errors and write to many digits. All not taking into account the possibility that something is very wrong.
Besides I am using here $c=1$ convention. Masses and energy are both measured in electronovolts, and speed is always compared to the speed of light, so this should not cause any problems.
