Let $P:C\to \mathbb{R}$, where $C$ is the Cantor set and $P$ is continuous. If $C_n$ is the sequence of numerators from the values of endpoints from the defined intervals of iteration $k$
$0, 1, 2, 3, 6, 7, 8, 9, 18, 19, 20, 21, 24, 25, 26, 27, 54, 55, 56, 57, 60, 61, 62, 63, 72, 73, 74, 75, 78, 79, 80, 81, 162, 163, 164, 165, 168, 169, 170, 171, 180, 181, 182, 183, 186, 187, 188, 189, 216, 217, 218, 219, 222, 223, 224, 225, 234, 235, 236, 237, 240, 241, 242, 243,...$
Where $C_1=0$, find the exact value of
$$\lim_{k\to\infty}\frac{\sum\limits_{n=1}^{2^{k+1}} P\left({C_n}/{3^k}\right)}{2^{k+1}}$$
For $P=x^2$ (According to Mathematica, the closed-form is $3/8$).
For $P={1}/{\left[2(x+2)\right]}$. $\left(\text{According to mathematica, it seems the closed form is} \left(\frac{1}{e}\right)^{-1-2e}\cot(e\pi)\cos(e\pi)\ \right)$
For $P=2^x$ (Doesn't appear to have a closed-form in fact most functions don't give a closed form.)
For $P$ in general (In terms of Integrals, Measure, etc.). This should solve my newly defined average for functions defined on the Cantor set.
Tl;dr: there is a recursive formula for monomials based on their degree given in a 1992 paper.
To start, I gathered some numerical evidence. For $0\le m,n \le 6$ I asked Mathematica to evaluate $\int_{C_m} x^n H^s(dx)$. For each fixed $n$, the partial sums had the formulas $$ \left\{1,\frac{1}{2},\frac{3^{-2 m}}{8}+\frac{3}{8},\frac{1}{16} 3^{1-2 m}+\frac{5}{16},\frac{9^{1-m}}{32}-\frac{17\ 9^{-2 m}}{320}+\frac{87}{320},-\frac{17\ 9^{-2 m}}{128}+\frac{25\ 9^{-m}}{64}+\frac{31}{128},-\frac{17}{512} 9^{1-2 m}+\frac{29\ 9^{1-m}}{512}+\frac{3253\ 9^{-3 m}}{46592}+\frac{10215}{46592}\right\}$$
Taking $\lim_{m\to \infty}$ you just keep the constant terms:
$$\left\{1,\frac{1}{2},\frac{3}{8},\frac{5}{16},\frac{87}{320},\frac{31}{128},\frac{10215}{46592} \right\}$$
I didn't recognize this sequence and it was somewhat intractable (RSolve, FindSequenceFunction, oeis were stumped). So I googled the sequence; amazingly I got a hit from a 1992 paper. Unfortunately, the paper was in Mandarin, of which I am wholly illiterate. So I asked a good friend to translate and she graciously agreed. Several months later, she emailed me a working draft. I'll give the main points from that paper and will link my edited version and the original (she asked that her version not be distributed).
One establishes that integrating with the Hausdorff measure is well-defined and agrees with the Riemann sum definition for continuous $f$.
Using a recursion formula for the right endpoints of each subinterval, one obtains the functional relation $$ \int _F f(x) H^s(dx ) = \int _F f(x) H^s(dx ) = \frac{1}{2}\left(\int _F f\left(\frac{x}{3}\right) H^s(dx )+\int _F f\left(\frac{x}{3}+\frac{2}{3}\right) H^s(dx) \right) $$
In particular, if we let $\alpha_n =\int_F x^n H^s(dx)$, with $\alpha_0=1$, then
$$\alpha_n = \frac{1}{2(3^n-1)}\sum_{k=1}^{n}\binom{n}{k}2^k\alpha_{n-k};$$
this is immediate from linearity and the binomial theorem. One checks that these values agree with the first few that we computed.
For continuous functions $g$ on $[0,1]$, one argues analogously: fix $\epsilon>0$ and use the Stone-Weierstrass theorem to approximate $g$ uniformly by a polynomial to within $\epsilon$. Then use the recursion to integrate the polynomial.
The translation was tricky at this point, but I think one can unite the Hausdorff measure with a 'singular' measure. The idea is to take a piecewise constant function on $[0,1]\setminus C_m$ and take the supremum to yield a function $h$ that is continuous with $h'(x)=0$ a.e. This induces a measure $\mu_h$, after which $$ \int _F f(x) H^s(dx) = \int _0^1 f(x) \mu_h(dx) $$
The rest of the paper talks about self-similar sets (of which the [ternary] Cantor set is one) and generalizes the similarity transformations to different parameters. In particular, there is a section on the von Koch curve that I have omitted.
I tried and failed to replicate some of the calculations you mentioned in your post. In particular, I'd love to see how you got the value of $\left(\frac{1}{e}\right)^{-1-2e}\cot(e\pi)\cos(e\pi)$ for that one function.