Thermodynamic Beta and Inverse Temperature

Since you are dealing with macrostates of constant energy, I assume you are dealing with microcanonical ensembles. The system is totally isolated and there is no exchange of heat, work or particles with the surroundings.

Suppose that the total energy of the system is E. Then:

f(E)=1/Ω(E)

where Ω(E) is the total number of microstates with energy E. Why is this true?

It is just the definition of the microcanonical ensemble wherein equal probability is assigned to all microstates of a given macrostate. Nothing surprising.

You then go on to cite a situation where all the energy belongs to one particle and all other particles have zero energy. This is not a macrostate, but a microstate where you have complete information about your system, there is no need for any statistical physics here.

Also from the reference in the linked question, the Boltzmann function is the probability distribution for individual energies in the macrostate, a higher probability for the lower energy particles and with an exponential decay. You should not be relating this with the probability of a microstate in the ensemble.

It would appear from the form of the Boltzmann distribution that for a system with fixed energy E the probability f(Ei) is always non-zero even when Ei>E (though it may be very small). Isn't this a problem?

Also the microcanonical ensemble assigns equal probability to all microstates that have the given energy E of the macrostate and zero probability to all other configurations. Hence there is no way any particle in any of the microstates will have an energy higher than E.

This seems like confusing notation. In a system of $N$ particles you don't normally talk about the energy of a single particle. A system of energy $E$ has some number of possible microstates, $Omega(E)$. The probability that the system, at any given moment, is in one particular microstate $r$ is (assuming the principle of a priori probability) $$P_r=frac{1}{Omega(E)}. tag{1}$$ This doesn't tell you anything about the energy of the individual particles. If you're talking about a specific microstate in which one particle has most of the energy that microstate still has a probability given by eq.(1) of existing at any given moment.

Secondly the entropy is usually defined as $$S=k_bln{Omega(E)}, tag{2}$$ at least where I've seen it, which is essentially just a measure of the number of microstates (and hence contains information about how likely you are to "guess" the exact microstate of a system at energy $E$ at any given moment).

Your argument from here loses me, it seems like you might have misunderstood the fundamentals. It's probably worth you revisiting them.

I am by no means an expert.

The connection between thermodynamic beta and inverse temperature can be established as follows (notation from Pathria, 3rd Ed.). Take two systems in thermal contact, each with $NVE$ macrostates (microcanonical ensembles). $$E^{(0)}=E_1+E_2tag{1}$$ We can see that the total number of microstates, $Omega^{(0)}$, is really just a function of either $E_1$ or $E_2$, $$Omega^{(0)}=Omega_1(E_1)Omega_2(E_2)=Omega_1(E_1)Omega_2(E^{(0)}-E_1)equivOmega^{(0)}(E_1) tag{2}$$ An axiom necessary in establishing equilibrium is that the total number of microstates of the combined system must be at a maximum, or in other words, $$frac{partial Omega^{(0)}}{partial E_1}=0 tag{3}$$ Therefore, $$frac{partial Omega^{(0)}}{partial E_1}=Omega_1frac{partial Omega_2}{partial E_1}+Omega_2 frac{partial Omega_1}{E_1}=0$$ $$frac{partial Omega_2}{partial E_1} = frac{partial Omega_2}{partial E_2}frac{partial E_2}{partial E_1}=-frac{partial Omega_2}{partial E_2}$$ $$frac{1}{Omega_1}frac{partial Omega_1}{partial E_1}=frac{1}{Omega_2}frac{partial Omega_2}{partial E_2}$$ $$frac{partial ln Omega_1}{partial E_1}=frac{partial ln Omega_2}{partial E_2} equiv beta_1=beta_2 tag{4}$$ Generalizing, $$beta = frac{partial ln Omega}{partial E} tag{5}$$ Where (4) is established by definition. We also know that, $$left ( frac{partial S}{partial E} right)_{N,V} = frac{1}{T} tag{6}$$ If we divide (6) by (5) and take non-differential changes, $$frac{Delta S}{Delta ln Omega}=frac{1}{beta T} = constant = k_B tag{7}$$ This is probably the most fundamental way to establish the relationship between $beta$ and $T$, but there are other links between them.

Edit: also worth noting that it can be shown that Gibbs' entropy generalizes to (7) when using $P_r = 1 / Omega$. So (7) really is the entropy formula (with no additive constant), but just for the microcanonical case.