You are asking a wrong question. Here is the problem with your reasoning.
You are assuming a Schwarzschild metric and a homogenous distribution of mass. But the Schwarzschild geometry describes a vacuum spacetime. So you can't use it for a spacetime filled with matter. For a cosmological spacetime filled with matter, like our universe, the suitable metric to use would be something else, like the FRW for example.
You could only use the Schwarzschild spacetime if you assumed a sphere of some uniform density $\rho$ and vacuum outside the radius of the sphere.
Let me illustrate how things would work out then. As you can see, a particular density corresponds to a particular $R_s$, lets call it $R_s(\rho)$. So if you had a sphere of matter with a radius $R_1$ grater than $R_s(\rho)$, then you couldn't apply the formula $R_s(\rho)=c\sqrt{\frac{3}{8\pi G \rho}}$. You would have to use the Schwarzschild metric only in the vacuum region outside of the sphere. So you would have then $R_s=\frac{8\pi G\rho R_1^3}{3c^2}$. In order to see how the $R_s$ compares with $R_s(\rho)$, you can replace the density with $\rho=\frac{3c^2}{8\pi G R_s(\rho)}$. So you would get that the Schwarzschild radius for a sphere of uniform density $\rho$ and radius $R_1>R_s(\rho)$ is $R_s=\left(\frac{R_1}{R_s(\rho)}\right)^2R_1$, which is grater than the radius of the sphere. So the sphere is inside its Schwarzschild horizon. If on the other hand, the radius $R_1$ is smaller than $R_s(\rho)$, then the corresponding horizon would have to be inside the sphere. But inside the sphere the Schwarzschild metric doesn't apply. So it isn't necessary that there should be a horizon inside the matter distribution.
If you apply these to the universe and assume for example that the radius of the visible universe is the radius $R_1$ of the sphere, then you would have a horizon radius (using your numbers) that would be almost 10 times the radius of the observable universe. So, the entire universe would have to be in a black hole of radius of 460 billion light-years. So the assumption that we should see black holes with horizons of radii of 13.9 billion light-years is not correct.
If one assumes the above point of view, one could say that the universe is a white hole that is exploding.
I hope that all these are helpful and not confusing.
One thing to note is that this horizon would only be present in an idealized eternal black hole, for a realistic non-rotating black hole formed by a collapsing star the Kruskal-Szekeres diagram would look more like the right-hand diagram below (from Gravitation by Misner, Thorne and Wheeler), where the gray area represents the interior of the star and the diagonal line labeled r=2M, t=infinity represents the sole event horizon:
In the case of an eternal black hole, then with the understanding that light rays always travel at 45-degree angles on a Kruskal-Szekeres diagram, you can see that while you can always see light rays from events on the antihorizon, you can never see light rays from events on the horizon until you cross it, and afterwards. You may find the animated gif from this page on Andrew Hamilton's site helpful:
Here red wavy lines represent worldlines of light rays (with the waviness just being symbolic, they should really be straight lines) from various events on the antihorizon, and blue wavy lines represent worldlines of light rays from events on the horizon.
Andrew hamilton also has some animated images showing falls into a Schwarzschild black hole, with a red grid representing the visible position of the antihorizon, and a white grid representing the visible portion of the horizon. See the "Through the horizon" section on this page for an animation with fancier graphics, and this page for a simpler version. It's explained a little more clearly on the second page--the red grid representing the antihorizon always appears off at some distance ahead of them even after they cross the horizon, but at the moment they cross they get their first light from events on the horizon, and that light appears as a white dot on the graphic but "The white dot is actually a line which extends from us to the Schwarzschild surface still ahead, though we only ever see it as a dot, not as a line". Then after they cross, this line expands into an white ovoid "bubble" which appears to intersects the red antihorizon, with the observer appearing to be enclosed within it:
The Schwarzschild surface appears to split into two in a region around
us. We seem to lie inside a bubble formed by the two surfaces. Looking
outward, we see the Schwarzschild surface (white grid lines) we just
passed through. If other observers fell through the surface after us,
we could see them fall through here, and they would appear neither
infinitely redshifted nor slowed.
Looking inward, we see the same Schwarzschild surface (red grid lines)
apparently still ahead. Persons who fell through before us would
continue to appear here. Already redshifted and slowed, such persons
would continue to grow ever more redshifted and slow.
Persons who appear to us to be inside the Schwarzschild bubble have
passed inside the horizon of the black hole. If they are sufficiently
close to us, then we can communicate with them, but they must be
close, for there's not much time left before we hit the central
singularity, not much time left for light signals to travel between
us.
Then the white bubble of the horizon continued to grow as the observer approaches the singularity.
So there is a sort of discontinuity in what you see as you cross the horizon, but in a way it's no different than the discontinuity when you cross any other light-like boundary like the surface of the future light cone of some event E (in fact the horizon and antihorizon are two sides of the future light cone of the event at the point where they cross). Prior to crossing an event's future light cone, you can't see light from events on the surface of that cone, but at the moment you cross you can see light from a single line of events going from E to your crossing-point, and after you've entered the cone you can see light from some spheroid of events in all directions away from you (the intersection of E's future light cone with your own past light cone).
Best Answer
This is a case of an unwisely chosen simile taken waaaay too far. This idea, that the entire universe could be inside the event horizon of not a supermassive, but rather a superduperultrahypermegastupendouslymassive black hole, is usually introduced in introductory classes about general relativity. The instructor in this case is trying to make clear that, contrary to a fairly popular misconception, the event horizon of a black hole is locally flat. That is, there are no CGI-fireworks, nor any kind of hard "surface", nor anything else particularly special, in the immediate vicinity of the event horizon. The only special thing that happens is a long distance effect, like noticing that every direction now points off in the distance towards the singularity.
The simile is also used to point out that, at the event horizon, even second-order, nearly local effects (that is, curvature of spacetime, or tidal effects in other words) become less pronounced the more massive the black hole is. (As an aside, this also explains why Hawking radiation is more intense for smaller black holes) So... as the simile suggests, if the black hole were massive enough, we might not even be able to detect it.
The key, though is massive enough. First of all, the whole beauty of the Einstein curvature tensor (the left side of Einstein's equation) is that it is Lorentz invariant, so it can be calculated in any reference frame, including one that is hypothetically based inside an event horizon.
The curvature can still be unambiguously calculated, so when you suggest that it may be only an optical illusion, you are also suggesting that all the scientists who do that type of large-scale curvature calculation (not me personally) are totally incompetent. Just so you know. I would suggest not mentioning that at any conferences on cosmology. One of the enduring mysteries of modern cosmology is that the large-scale curvature of the Universe seems to be open (like the 3-space-plus-one-time dimensional analog of a saddle or Pringle potato chip) and not flat (like Euclidean geometry) or closed (like a sphere). The last is what we would calculate if the visible Universe were inside a black hole.
So, for the visible Universe to be inside an event horizon, the Cosmic Acceleration we have seen thus far would have to actually just be one small, contrarian region inside an even larger event horizon of globally closed curvature. Just to make the event horizon radius 13.7 giga-lightyears (a bare minimum starting point that excludes all manner of things that make the real situation many orders of magnitude worse*), you would need over 8E52 kg of mass in the singularity. This would require over 5E79 protons, where I have heard that the entire visible Universe only has about 10^80 particles, total, and I think I heard that there are about 10^18 photons for every proton, or maybe even all other particles. Somebody can look that up if they want to, but it's definitely a big number. The upshot is that there would have to be an amount of mass, all crammed into one singularity, that would render the total mass of every single thing we can see a barely detectable rounding error. Monkeying with all those dark matter and even dark energy theories is less of a leap than that.
Your prediction doesn't actually predict anything, since you account for either its presence or its absence.
Speculation 1: Olber's Paradox is already solved for accepted theories of cosmology, so pointing out that your theory can also resolve it is nice but doesn't score any points.
Speculation 2: Are you suggesting that the singularity is where all the antimatter to match the Universe's matter went? Remember the singularity dwarfs the visible Universe. That still doesn't explain the asymmetry, it only pushes the question back by one logical step: Why did the antimatter go into the big singularity and not the matter?
Speculation 3: Hawking radiation for the big singularity's event horizon lends whole new meaning to the term negligible. See my previous aside. Also, we can't observe matter being destroyed at the singularity. That's information flowing the wrong way. Also, that negates the previous assertion that the sky is black because it's towards the singularity.
*Like cosmic expansion, just how small our contrarian region is, compared to the whole event horizon, and probably some other, subtler things.