I'm wondering if there exists such a function, whose $(n+1)$th derivative is defined only on a proper subset of the domain where the nth derivative is defined, and with the property that the diameter of the nth domain tends to zero as $n$ approaches infinity and therefore at and only at $1$ single point this function is smooth.
A motivating example is the integral of the function
$$f(x)= \begin{cases} 1 & : x \in [0,1]\\ \textrm{Weierstrass function }+c_1 & : x < 0 \\ \textrm{Weierstrass function }+c_2 & :x >1 \end{cases} $$
$c_1$ and $c_2$ are chosen such that $f(x)$ is continuous.
Additional question: what if the function is defined over $\mathbb{C}$?
Yes, there is such a thing. First, construct the sequence of functions $f_n:\mathbb{R} \to\mathbb{R} $, $n\ge 1$ such that
For example, one can construct $f_n$ by taking the indefinite integral of the Weierstrass function $n$ times, and multiplying the result by the bump function $$ \phi(x) = \begin{cases} \exp((x-1)^{-1}(x-3)^{-1}),\quad & 1<x<3 \\ 0 & \text{otherwise}\end{cases} $$ Separately, define a continuous $f_0$ that is nowhere differentiable on $(1,\infty)$ (e.g., the Weierstrass function plus a constant) and is zero on $(-\infty,-1)$.
Then the function $$f(x) = \sum_{n=1}^\infty \exp(-2^n) f_n(2^n|x|)\tag{1}$$ has the desired properties. Indeed, $f$ is differentiable on $(-1,1)$ but not on any larger open interval. Its derivative $f'$ is differentiable on $(-1/2,1/2)$ but not on any larger open interval. And so forth: every point $x\ne 0$ has a neighborhood in which all terms of $(1)$ vanish except at most two; and of those two one is exactly $n$ times differentiable while the other is $n+1$ times differentiable. So the sum of them two is exactly $n$ times differentiable.
At $0$, all derivatives of $f$ turn to $0$. Indeed, for every $k$ the series $\sum_{n=1}^\infty \exp(-2^n) \frac{d^k}{dx^k}f_n(2^n|x|)$ converges uniformly in some neighborhood of $k$, which implies $k$th order differentiability of $f$ there.