The cross-section for neutrino interactions is energy dependent.
For solar neutrinos at $\sim 0.4$ MeV, which would likely dominate any neutrinos likely to interact with a brain (the cosmic background neutrinos have way low energies), the cross-sections are $\sigma \sim 10^{-48}$ m$^2$, for both leptonic processes (elastic scattering from electrons) and neutrino-nucleon interactions.
The mean free path of a neutrino will be given by $l \sim (n\sigma)^{-1}$, where $n$ is number of interacting target particles per cubic metre and $\sigma$ is the cross-section.
If your head is basically water with a density of 1000 kg/m$^3$, then there are $n_e = 3.3\times10^{29}\ m^{-3}$ of electrons and about $6 \times 10^{29} m^{-3}$ of nucleons.
Including both nucleonic and leptonic processes, the mean free path is $\sim 10^{18}\ m$.
So unless your head is 100 light years wide, there is little chance of any individual neutrino interacting with it.
This is only one part of the calculation though - we need to know how many neutrinos are passing through your head per second. The neutrino flux from the Sun is about $7\times 10^{14}$ m$^{-2}$ s$^{-1}$. If your head has an area of about 400 cm$^2$, then there are $3\times 10^{13}$ neutrinos zipping through your brain every second.
Thus is we take $x=20$ cm as the path length through your head, there is a chance $\sim x/l$ of any neutrino interacting, where $l$ was the mean free path calculated earlier.
This probability multiplied by the neutrino flux through your head indicates there are $6\times 10^{-6}$ s$^{-1}$ neutrino interactions in your head, or roughly one every two days.
Whether that would produce any perceptible effect in your brain needs to be shunted back to Biology SE. If we require it (or rather scattered electrons) to produce Cherenkov radiation in the eyeball, then this needs $>5$ MeV neutrinos and so the rate would reduce to 1 per 100 days or even lower due to the smaller number of neutrinos at these energies and the smaller volume of water in the eyeball.
EDIT:
In fact my original answer may be over-optimistic by an order of magnitude since water only acts as a good detector (via Cherenkov radiation) for neutrinos above energies of 5 MeV. Solar neutrinos are predominantly lower energy than this. My calculation ignored atmospheric neutrinos which are produced in far fewer numbers (but at higher energies $\sim 0.1-10$ GeV). The cross-section for these is 4-6 orders of magnitude higher, but I think they are produced in so much lower numbers that they don't contribute.
Conclusion It doesn't have anything to do with neutrinos. The rate would be too low, even if they could be perceived.
