Heterozygosity and overdominance
If the fitness of a heterozygote is $(1+s/2)$ and of a homozygote is $(1-s/2)$ then why is the probability for a given state $(1+s/2)^j(1-s/2)^{m-k}$

$$\binom mj (1/2)^j(1/2)^{m-j}= \binom mj (1/2)^m~~ ?$$

As you pointed out earlier, in the general case it need not be true that $p = q = 1/2$ but that is what the form of the probability above implies. So the threshold question is why this particular heterozygous-dominant model implies equilibrium probabilities $p = q = 1/2.$ I think the ideas below begin to address this.

The simplest case is for one locus, two alleles, and there are many good derivations online. I think if you understand the situation for one locus you can generalize to higher numbers. (Hopefully I will supplement this answer, time allowing. I think a shortcut would be to assume $p = q = 1/2$ and use your fitness weights. That gives us $\bar{w}=1$ as a denominator. Now for reasons of symmetry I think you can show that the relative frequencies of $p$ and $q$ are equal and so $p' = p.$ Then you have to show that this equilibrium solution is unique.) 

For a single locus the derivations of `heterozygote advantage' I have found,(1)(2), assign fitness weights as follows:

AA = (1 - s),     Aa = 1,       aa = (1 - t )

in which $s,t &gt; 0, s \neq t$ in general, from which they derive as a condition for equilibrium 

$$\Delta q = \frac{pq(sp-tq)}{\bar{W}}  = 0$$

so 

$$\hat{p} = \frac{t}{s+t} ~~~\text{and}~~~ \hat{q} = \frac{s }{s+t}$$

in which $\hat{p}, \hat{q}$ are equilibrum frequencies for each allele, respectively, and s and t are rates of mutation. Nowhere did I see a model in which they had assigned the same fitness to both homozygous cases (AA, aa) but it's just a special case of the heterozygote advantage model. 

fitness (AA) = fitness(aa) = (1 - s/2), fitness(Aa) = (1+ s/2). 

So if we subtract s/2 from each fitness score (or normalize with respect to Aa to the same effect), we get: 

fitness(AA) = fitness (aa) = (1- s ) and fitness (Aa) = 1 as in the two references above, except that now the two homozygous states have equal fitness. 

But then we have $$\Delta q = \frac{pq(sp-sq)}{\bar{W}} =  \frac{pqs(p - q)}{\bar{W}} $$

and the only nontrivial solution is $p = q = 1/2.$ 

So what I am suggesting is that when you assign the same fitness to both $AA$ and $aa$ you no longer have the general case. As for the value of s being forced, the following seems relevant.

The full expression from (2) for the equilibrium condition is 

$$p' = \frac{p^2 W_{AA}+ pq W_{Aa}}{\bar{W}} \hspace{10mm}(1)$$

in which$ W_{AA} = W_{aa} =  1- s$ and $W_{Aa} = 1$ and 

$\bar{W} = p^2 W_{AA}+2pqW_{Aa}+q^2W_{aa}$

If the homozygotes are assigned the same fitness and $p = q = 1/2$ , equation (1) above becomes:

$$p' = \frac{\frac{1}{4} + \frac{1}{4}(1-s)}{\frac{1}{2} + \frac{1}{2}(1-s)} $$

The program on page 583 of (1) is helpful. Let h:= heterozygous and m:= homozygous. If $p = q = 1/2$ then as long as fitness(h) and fitness(m) are equal the system is in equilibrium immediately. 

If $p\neq 1/2$ then as long as f(h) = f(m) the system reaches equilibrium at $p = 1/2$ asymtotically. If $p = 1/2$ but f(h) $\neq$ f(m) the asymtotic limit has to be calculated.  

See also http://evol.bio.lmu.de/_teaching/evogen/Evo8-Summary.pdf