Since the excised intervals vary in their length while the remaining intervals are not, it seems easier to focus on the remaining intervals.
Let $I(n,k)$ for $n \geq 1$ and $0 \leq k \leq 2^n - 1$ be the remaining $2^n$ intervals after the $n$-th stage of the construction of the Cantor set. Then $|I(n, k)| = 3^{-n}$, and we can approximate $f$ by
$f_n(x) = \int_{0}^{x} \left( (3/2)^n \sum_{k=0}^{2^n - 1} \chi_{I(n,k)}(t) \right) \, dt $
To see this really approximates $f$, observe that $f_n$ increases only on $C(n) = \bigcup_{k=0}^{2^n-1}I(n,k)$ and on each subinterval $I(n,k)$, $f_n$ increases by exactly $2^{-n}$, as we can check:
$ \int_{I(n,k)} (3/2)^n \chi_{I(n,k)}(t) \, dt = \frac{1}{2^n}.$
Thus $f_n$ coincides exactly with the $n$-th intermediate function appearing in the construction of the Cantor-Lebesgue function $f$. Then $f_n \to f$ uniformly, and we have
$ \begin{align*} \int f(t) \, e^{-ixt} \, dt &= \lim_{n\to\infty} \int f_n(t) \, e^{-ixt} \, dt \\ &= \lim_{n\to\infty} \left( \left[ -\frac{1}{ix} f_n(t) e^{-ixt} \right]_{0}^{1} + \frac{1}{ix} \int f_n'(t) \, e^{-ixt} \, dt \right) \\ &= -\frac{e^{-ix}}{ix} + \frac{1}{ix} \lim_{n\to\infty} \int f_n'(t) \, e^{-ixt} \, dt \\ &= -\frac{e^{-ix}}{ix} + \frac{1}{ix} \left(\frac{3}{2}\right)^n \lim_{n\to\infty} \sum_{k=0}^{2^n - 1} \int_{I(n,k)} e^{-ixt} \, dt \end{align*}$
Now, direct calculation shows that
$\int_{a}^{a+\beta h} e^{-ixt} \, dt + \int_{a+(1-\beta)h}^{a+ h} e^{-ixt} \, dt = 2 \cos\left(\frac{1-\beta}{2} hx\right)\frac{\sin(\frac{\beta}{2}hx)}{\sin(\frac{1}{2}hx)} \int_{a}^{a+h} e^{-ikt} \, dt. $
Thus plugging $h = 3^{-n}$ and $\beta = \frac{1}{3}$, we have
$\int_{I(n+1,2k)} e^{-ixt} \, dt + \int_{I(n+1,2k+1)} e^{-ixt} \, dt = 2 \cos\left(\frac{x}{3^{n+1}}\right)\frac{\sin\left(\frac{x}{2\cdot 3^{n+1}}\right)}{\sin\left(\frac{x}{2\cdot 3^{n}}\right)} \int_{I(n,k)} e^{-ikt} \, dt. $
Inductively applying this relation allows us to calculate the limit above, which I leave because I have to go out.