<s>[INST] About the "Cantor volume" of the $n$-dimensional unit ball A simple derivation for the Lebesgue measure of the euclidean unit ball in $\mathbb{R}^n$ follows from computing
$$ \int_{\mathbb{R}^n}e^{-\|x\|^2}\,dx $$
in two different ways. See, for instance, Keith Ball, An Elementary Introduction to Modern Convex Geometry, page $5$. Now I was wondering about the following slightly unusual variation:

Let $X_1,\ldots,X_n$ independent random variables with the Cantor
  distribution. 
What is the probability that $X_1^2+\ldots+X_n^2\leq 1$?

I bet this can be tackled by exploiting the fact that the cumulants of the Cantor distribution are given by:
$$ \kappa_{2n}=\frac{2^{2n-1}(2^{2n}-1)\,B_{2n}}{n (3^{2n}-1)},\tag{CM}$$
but how to prove $(\mathrm{CM})$? - This has been answered, but the main question is still open.
 [/INST] HINT:
For $\mu(x)$ the Cantor measure supported on the Cantor set $\subset [0,1]$ we have the change of variable formula:
$$\int f(x)\, d\mu(x) = \frac{1}{2} \int f(1/3 x)\, d\mu(x)  + \frac{1}{2} \int f(1/3 x  + 2/3)\, d \mu(x)$$
analogous to $\int_0^1 f(x)\, dx =\frac{1}{2} \int_0^1 f(1/2 x)\, d x + \frac{1}{2} \int_0^1 f(1/2 x + 1/2)\, d x $
$\bf{Added:}$ 
It's easy to see that the first moment $E(X) = \int x \, d \mu(x)= \frac{1}{2}$, and this can be obtained readily from the above formula for $f(x) = x$. 
Consider now the central moment generating function 
$$F(t)\colon =E[e^{t(X-\frac{1}{2})}] = \int e^ {t(x-\frac{1}{2})} \, d\mu(x)                     $$
From the above equality for $f_t(x) = e^{t(x-\frac{1}{2})}$ we get 
$$\int e^ {t(x-\frac{1}{2})} \, d\mu(x)  = \frac{1}{2}\left( \int e^ {t(\frac{x}{3}-\frac{1}{2})} \, d\mu(x) + \int e^ {t(\frac{x}{3}+\frac{2}{3}-\frac{1}{2})} \, d\mu(x) \right) $$
Now we notice that 
\begin{eqnarray}
t\,(\frac{x}{3}-\frac{1}{2})= \frac{t}{3}(x - \frac{1}{2}) - \frac{t}{3}\\
t\,(\frac{x}{3}+\frac{2}{3}-\frac{1}{2})= \frac{t}{3}(x - \frac{1}{2}) + \frac{t}{3}
\end{eqnarray}
Therefore we get the equality
$$F(t) = \frac{e^{\frac{t}{3}} + e^{-\frac{t}{3}}}{2} \cdot F(\frac{t}{3}) 
$$
$\bf{Added:}$ Rewrite the above equality as 
$$F(3t) = \frac{e^t + e^{-t}}{2} F(t)$$
Let $G(t) = \log F(t)$. From the above we get
$$G(3 t) - G(t) = \log ( \frac{e^t + e^{-t}}{2})$$
$\bf{Added:}$ 
Some (moment) calculations: 
$$\int x d \mu(x) = \frac{1}{2} \left( \    \int (\frac{1}{3} x + \frac{1}{3} x + \frac{2}{3}) d\mu(x) \right )$$
implies $\int x d \mu(x) = \frac{1}{2}$ as expected.
Let's apply the same formula for $f(x) = (x-\frac{1}{2})^n$. We have
\begin{eqnarray}
\int (x-\frac{1}{2})^n d \mu(x) = \frac{1}{2}\left( \int ( \frac{x}{3} - \frac{1}{2})^n + ( \frac{x}{3} + \frac{2}{3}- \frac{1}{2})^n d\mu(x) \right ) = \\
=\frac{1}{2\cdot 3^n}\left( \int ( x - \frac{1}{2}-1)^n + ( x-\frac{1}{2} + 1)^n d\mu(x) \right )
\end{eqnarray}
that is
$$M_n = \frac{1}{3^n}\sum_{k \ge 0} \binom{n}{2k} M_{n-2k}$$
which is basically a formula from above $F(3t) = \frac{e^t + e^{-t}}{2} F(t)$.
We get from here $m_2 = \frac{1}{8}$, $m_4 = \frac{7}{320}$, etc. Note that the formula provided in Wikipedia is for the cumulants, not the central moments, as $\kappa_4 = \frac{1}{40}$.
</s>