The answer to this question is addressed, for the case of a simple Schwarzschild black hole, by Taylor & Wheeler in their book called "Exploring black holes" (Addison, Wesley, Longman, 2000, pages B20-B24). There is a huge difference to what would be perceived by a "shell" observer that arranges to be stationary just outside the event horizon compared with an observer that is free-falling into the black hole.
Your question supposes that an observer is falling into the black hole. That observer will be able to see the outside universe from inside the event horizon. The light will be blue-shifted and it will be distorted/aberrated in such a way that as the observer approaches the singularity, the light from the rest of the universe is pushed outward (i.e. the viewing angle with respect to the fall-line becomes larger) into a halo and finally an intense ring of blue-shifted radiation that goes all around the sky, with blackness both in front of the observer and behind. Nothing special happens to the observer as they cross the event horizon.
A "shell" observer, using almost-infinite rocket power to hover just above the event horizon would see the whole universe compressed to a small, intense, blue-shifted dot overhead.
Of course there cannot be "shell" observers inside the event horizon since everything is compelled to move towards the singularity.
NB: This all just assumes classical GR theory. For further information and animations you could look at Andrew Hamilton's set of resources.
It is not the cosmic censorship theorem/conjecture that rules out stable toroidal black holes but no-hair theorem/conjecture. If such black hole existed its field expansion would have contained additional terms (anapole moment ?) which would have constituted 'hair' in terms of the theorem.
For instance, in higher dimensions there is 'cosmic censorship' conjecture but no 'no-hair' theorem. And there are 'black ring' solutions: in 5D it hash $S_2\times S_1$ horizon topology (which would be the analogue of toroidal black hole).
Best Answer
I think it's fair to say that the EHT image definitely is consistent with GR, and so GR continues to agree with experimental data so far. The leading paper in the 10th April 2019 issue of Astrophysical Journal letters says (first sentence of the 'Discussion' section):
I'm unhappy about the notion that this 'confirms' GR: it would be more correct to say that GR has not been shown to be wrong by this observation: nothing can definitively confirm a theory, which can only be shown to agree with experimental data so far.
This depends of course on the definition of 'confirm': above I am taking it to mean 'shown to be correct' which I think is the everyday usage and the one implied in your question, and it's that meaning I object to. In particular it is clearly not the case that this shows 'Einstein was right': it shows that GR agrees with experiment (extremely well!) so far, and this and LIGO both show (or are showing) that GR agrees with experiment in regions where the gravitational field is strong.
(Note that, when used informally by scientists, 'confirm' very often means exactly 'shown to agree with experiment so far' and in that sense GR has been confirmed (again) by this observation. I'm assuming that this is not the meaning you meant however.)
At least one other answer to this question is excellent and very much worth reading in addition to this.