Let's say I was at the very center of the enormous Boötes void, way out in deep, deep space. What could I see with the naked eye? I assume I could see no individual stars, but could I resolve any galaxies? If I gazed in the direction of a super-cluster of galaxies would it seem brighter than other directions? How dark would it be compared to, say, the far side of the moon when it is a full moon on earth?
I am told there are, in fact, a few galaxies in the void. So let's say I pick a spot in the void that is as far from any of those galaxies as possible.
Best Answer
Individual sources
The number density of galaxies in a void is typically an order of magnitude lower than the average in the Universe (e.g. Patiri et al. 2006). In this astronomy.SE post, I estimate the number density of galaxies of magnitude $M=-17$ or brighter in the Boötes Void to be $n \sim 0.004\,\mathrm{Mpc}^{-3}$, or $10^{-4}\,\mathrm{Mlyr}^{-3}$ (i.e. "per cubic mega-light-year"). Hence, the typical distance to a galaxy from a random point in the Boötes Void is $$ d = \left( \frac{3}{4\pi n} \right)^{1/3} \simeq 13\,\mathrm{Mlyr}. $$
Although some galaxies will be brighter than $M=-17$, the number density declines fast with brightness; for instance, galaxies that are 10 times brighter are roughly 100 times rarer, meaning that they're on average 5 times more distant and hence appear 25 times fainter. On the other hand, the number density of galaxies fainter than $M=-17$ doesn't increase that fast (in astronomish: $-17$ is close to $M^*$; "M-star").
So for the sake of this calculation, let's assume that the closest galaxy is an $M=-17$ galaxy at a distance of $13\,\mathrm{Mlyr}$. That distance corresponds to a distance modulus of $\mu \simeq 28$, so the apparent magnitude of the galaxy would be $$ m = M + \mu \simeq 11. $$
Typically, humans cannot see objects darker than $m \simeq 6.5$ (the magnitude scale is backwards, so darker means "larger values than 6.5"), although some have claimed to be able to see $m\simeq8$ — still an order of magnitude brighter than the $m=11$ estimated above. Moreover, this threshold assumes point sources, whereas a galaxy has its brightness smeared out over a quite large area, lowering its surface brightness significantly!$^\dagger$. Note also that, as in the rest of the Universe, galaxies in voids are not completely randomly scattered throughout space, but tend to cluster in clumps and filaments, and that the number density is smaller in the center of the void, meaning that here the typical distance to the next galaxy is larger.
Hence, you would — at a random position in the Böoted Void — be most likely to be floating in complete darkness.$^\ddagger$
Background radiation
The combined light from all astrophysical and cosmological sources comprises a cosmic background radiation (CBR), meaning that at any time your eye does indeed receive photons across the entire electromagnetic spectrum. Thus the term "complete darkness" may be debated. On average, this background is dominated by the cosmic microwave background (if you're close to a star or a galaxy, those sources will dominate, but then it isn't really a "background" any longer).
In this answer, I estimate the total background in the visible region (from sources outside the Milky Way) to be roughly $3.6\times10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}$. If I've done my maths right, this corresponds to a the light from a 25 W light bulb, smeared out over a 15 km diameter sphere with you in the center. The Böotes Void would have an even lower background than this. I'm not a physiologist, but I think this qualifies as "complete darkness" (to the human eye; not to a telescope).
$^\dagger$For instance, the Andromeda galaxy has an apparent magnitude of $m=3.44$ which, if its light were concentrated in a point, would make it easily visibly even under light-polluted conditions.
$^\ddagger$Your eye might be able to detect individual photons, at stated in Árpád Szendrei's answer, but that hardly counts as "seeing anything".