I think the following portion from your question is most important:
But, the halting-detection machines used by the Busy Beaver researchers don't have too much power. They have too little power. Currently they can't solve $n=5$. OK, so we give them some more power. Maybe at some point they can solve $n=5$ ... but they still can't solve $n=6$. Maybe we can give them enough power to solve $n=6$, or $n=7$
or ....
... so my question is: is there some 'special' value of $n$, say $n=m$ where this has to stop. Where, somehow, the only way to solve $n=m$, is by a machine that has 'too much' power? But why would that be?
The solution to solving $\Sigma(5)$ isn't simply giving Turing machines "more power." The reason we don't know $\Sigma(5)$ right now is because there are 5-state Turing machines which mathematicians believe will never halt, but have not been able to prove will never halt. The problem is not as simple as just enumerating through all of the 5-state Turing machines since once you've done that, you still need to figure out if the Turing machine halts or not, which, as you know, is not a trivial problem. We've been able to do this for 4-state Turing machines, but we don't know yet if we can do this for 5-state Turing machines because there may very well be 5-state Turing machines which we can never prove to be halting/non-halting within the system of classical mathematics (that is, ZFC).
Now, you've asked what is the magic number is: What is the magic number $n=m$ such that we will never be able to solve for $\Sigma(n)$? As I've said above, that magic number could very well be $n=5$, but that hasn't been proven yet. However, mathematicians have proven that $\Sigma(1919)$ is indeed un-knowable within ZFC. Before explaining this, I think it might be helpful to quote your question again:
Then again, these machine don't do any explicit analysis of the machines involved ... they just happen to give the right value. So, maybe there is still some value of $n$ where the approach of actually trying to analyze and predict the behavior of machine starts to break down for some fundamental, again possibly self-referential, reason?
My answer to this question is yes, there is a 1919-state Turing machine such that trying to predict if the machine halts would be fundamentally unsolvable within our system of mathematics. See, the way mathematicians proved $\Sigma(1919)$ is unsolvable is by constructing a 1919-state Turing machine $M$ which halts if there is a contradiction within ZFC and never halts if ZFC is consistent. However, there's no way to prove ZFC is consistent using the axioms of ZFC because of Godel's Second Incompleteness Theorems. This means we can never use the ZFC axioms of mathematics to prove machine $M$ ever halts or not because doing so would constitute a proof that ZFC is consistent. Thus, mathematicians can't predict if machine $M$ halts or not because of Godel's Incompleteness Theorem, which means that the busy-beaver problem for 1919-state Turing machines is unsolvable.
I hope this gives you some more insight into why $\Sigma(n)$ is solvable for small values of $n$ but unsolvable for larger values of $n$. Anyway, I am certainly not an expert in theory of computation, so if someone would like to add an alternate explanation/clarifying comments to my answer, feel free. Thanks!
With your approach, the assumption on will_halt that you start out with "Assume we had a program will_halt(source, args) that returns True if eval(source, args) halts and False otherwise." does not hold anymore: there are cases where evaluating source runs forever, but your will_halt returns True. So, whatever you do afterwards does not address the halting problem anymore, because that starts out with your (original) assumptions.
(As an aside, note that it is trivial to write a will_halt function that returns True if the eval(source,args) terminates and without restriction on the return value if eval(source,args) does not terminate. It's useless, though.)
Best Answer
Yes, we can assume there is a "universal halting predictor" $H$ that produces output $0$ if Turing machine $P$ with input $I$ never halts and $1$ if its does halt - and we assume $H$ always outputs either $0$ or $1$ and is always correct.
But the point of the proof is that we could then construct a Turing machine $H'$ which, when given input $I$, feeds its own description together with input $I$ to $H$. $H'$ then does the opposite of whatever $H$ predicts it will do. This means that $H$ cannot predict what $H'$ will do with input $I$, so $H$ is not after all a universal halting predictor for Turing machines.
Note that this argument does not rule out a "magic" (i.e. non-algorithmic) halting predictor $M$ for Turing machines. It would not be possible to construct a contradictory $M'$ around $M$ because the non-algorithmic $M$ is part of $M'$, so $M'$ does not have an algorithmic description which it could feed to $M$. In other words, magic $M$ may be able to predict whether any Turing machine will halt, but $M$ and $M'$ are not Turing machines.