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!
For any machine $M$ on any input $x$, you can (using snm) prove that there is a machine $M_x$ that do the same compution than $M$ on input $x$, on any input :
$$\forall M\forall x \exists M_x\forall y \quad M(x)=M_x(y)$$
Then, all the information is in the diagonal, because $M_x$ on its own description compute also $M(x)$.
Best Answer
Yes.
It will be convenient to introduce some jargon and notation. The notion of computing from an oracle gives rise to the preorder of Turing reducibility (or "relative computability," or similar), which we denote "$\le_T$." This in turn gives rise to an equivalence relation on decision problems: $A\equiv_T B$ if $A$ computes $B$ and $B$ computes $A$. The resulting partial order of $\equiv_T$-classes is called the (poset of) Turing degrees, and is generally denoted by "$\mathscr{D}$" or similar.
The properties of this partial order have been extensively studied. Some results include:
Given any nonzero (= not the degree of the computable sets) Turing degree $\bf d$, there is a $\bf\hat{d}$ which is incomparable with $\bf d$ (this answers your question).
In fact, any (nonzero) Turing degree is contained in an antichain of degrees of size continuum.
Every Turing degree is above only countably many degrees, so the "height" of the Turing degrees is $\omega_1$; in case $\mathsf{CH}$ fails, this means the poset of Turing degrees are "wider" than it is "tall" (and even if $\mathsf{CH}$ holds this is still a useful heuristic in certain ways).
The Turing degrees form an upper semilattice - given decision problems $A, B\subseteq\omega$, the join of their degrees is the degree of $\{2n: n\in A\}\cup\{2n+1: n\in B\}$. Moreover, this semilattice has no top element (because of the "relativized" halting problem, or Turing jump).
However, their exist Turing degrees $\bf d$, $\bf \hat{d}$ with no greatest lower bound. So $\mathscr{D}$ is not a lattice.
Moreover, nontrivial infinite joins never exist (Exact Pair Theorem): given an infinite set of degrees $D$, if $D$ is nontrivial (= does not have a finite subset $D_0$ such that $\forall {\bf d}\in D\exists {\bf d_0}\in D_0({\bf d}\le_T {\bf d_0})$) and ${\bf a}$ is an upper bound of $D$, then there is a degree ${\bf b}$ which also is an upper bound of $D$, and which is incomparable with ${\bf a}$; moreover, $\{{\bf d}: {\bf d}\le_T {\bf a}$ and ${\bf d}\le_T {\bf b}\}=D$.
If $A'$ and $B'$ are the halting problems of $A$ and $B$, and $A\le_T B$, then $A'\le_T B'$. This means the jump can be thought of as an operation on degrees, not just sets. It turns out this operation is definable just in terms of the partial ordering! This was an extremely surprising result; see these notes of Slaman.
The converse of the above bullet point fails extremely badly: the jump is extremely non-injective. For example, for every ${\bf d}\ge_T{\bf 0'}$ there is a ${\bf c}$ such that ${\bf c'}={\bf d}$, ${\bf c}>_T{\bf 0}$, and there is no degree strictly between ${\bf c}$ and ${\bf 0}$ (such degrees are called "minimal," with the more exact phrase "minimal nonzero" being a bit long to say).
The above is all about the global theory of the Turing degrees; people have also studied extensively the local theory of special subclasses of Turing degrees. The best known such class is the class of degrees of domains of partial computable functions, the c.e. degrees. Of course, the degree of the halting problem is the maximal c.e. degree, so in this context the answer to your question is “no;” however, given any c.e. degree which is not the degree of the halting problem or the computable sets, there is an incomparable c.e. degree. This is the Friedberg-Muchnik Theorem, and its proof was a precursor to the method of forcing in set theory.
Let me end by stating my favorite two open questions about the Turing degrees:
Does the poset of all Turing degrees have any nontrivial automorphisms?
Does the poset of c.e. degrees have any nontrivial automorphisms?
Back in the day both these degree structures were believed to be very homogeneous, with lots of automorphisms; the theme of pure computability theory post-1955ish, however, was quite the opposite: the Turing/c.e. degrees are structurally rich, with lots of definable subclasses, and in fact it is now believed that both partial orders are rigid. All that is currently known however is that the automorphism groups are at most countable.
Finally, having said that the answer to your question is “yes,” let me explain a sense in which the answer to your question is “no.” The only “natural” increasing functions on the Turing degrees which have been discovered so far are essentially iterates of the Turing jump. Martin conjectured that this holds for "reasonable" functions. Ignoring some subtleties around the notion of iteration here there are a few ways to make this conjecture precise, the two most common being the following:
(1) $\mathsf{ZFC}$ should prove that every Borel function which is degree-invariant and increasing is an iterate of the jump, at least when we restrict to some set of the form $\{{\bf d}:{\bf d}\ge_T{\bf c}\}$ (that is, "on a cone").
(2) $\mathsf{ZF+DC+AD}$ should prove that every increasing function whatsoever is an iterate of the jump on a cone - the point being that under $\mathsf{ZF+DC+AD}$ all functions are reasonably nice.
Currently weak versions of Martin’s conjecture have been proved, although the full conjecture is still very much open.