For a $p$-adic number $a \in \frac{n}{p^k}+\mathbb{Z}_p\subset \mathbb{Q}_p$ let $\exp(2i \pi a) =\exp(2i \pi \frac{n}{p^k})$ and $\psi_a(x) = \exp(2i \pi ax)$. Then $$Hom(\mathbb{Z}_p,\mathbb{C}^\times) = \{\ \psi_a , a \in \mathbb{Q}_p/\mathbb{Z}_p\}$$
For $f,h$ (uniformly) continuous $ \mathbb{Z}_p \to \mathbb{C}$ let $\langle f,h \rangle =\lim_{k \to \infty} p^{-k} \sum_{n=0}^{p^k -1} f(n) \overline{h(n)}$
$f \ast h(x) = \langle f(x-.),\overline{h} \rangle $
We obtain that $$\lim_{k \to \infty} \sum_{n=0}^{p^k -1} \langle f,\psi_{n/p^k}\rangle \psi_{n/p^k}(x) = \lim_{k \to \infty} f \ast \sum_{n=0}^{p^k -1} \psi_{n/p^k}(x)= \lim_{k \to \infty} f \ast p^k 1_{x \in p^k \mathbb{Z}_p} = f(x)$$ and the convergence is uniform.
This is to be interpreted as a particular order of summation for the Fourier series $$\sum_{a \in \mathbb{Q}_p/\mathbb{Z}_p}\langle f,\psi_a\rangle \psi_a(x) \tag{1}$$
Question : when does $(1)$ converge absolutely ?
If $f$ is locally constant, that is for some $m$, $f(x) = f(x \bmod p^m)$ then its Fourier series is a finite sum $\sum_{a \in p^{-m}\mathbb{Z}_p/\mathbb{Z}_p}\langle f,\psi_a\rangle \psi_a(x)$ as $\langle f, \psi_{n/p^k} \rangle = 0$ whenever $ k >m$ ($p\nmid n$).
With $\|f\|_2^2 = \langle f,f \rangle$ we have the Hilbert space $L^2(\mathbb{Z}_p)$ of which $\{\ \psi_a , a \in \mathbb{Q}_p/\mathbb{Z}_p\}$ is an orthonormal basis so $f \in C^0(\mathbb{Z}_p) \implies f \in L^2(\mathbb{Z}_p) \implies \sum_{a \in \mathbb{Q}_p/\mathbb{Z}_p}|\langle f,\psi_a\rangle|^2 = \|f\|_2^2$. Unlike the derivative $L^2(\mathbb{R}/\mathbb{Z})\to L^2(\mathbb{R}/\mathbb{Z})$ there is no obvious unbounded operator such that $T(f) = \sum_{a \in \mathbb{Q}_p/\mathbb{Z}_p}c(a) \langle f,\psi_a\rangle \psi_a$ and $\sum_{a \in \mathbb{Q}_p/\mathbb{Z}_p}\frac{1}{|c(a)|^2} < \infty$. If that operator could be easily defined we'd have $T(f) \in L^2(\mathbb{Z}_p) \implies $ $$\sum_{a \in \mathbb{Q}_p/\mathbb{Z}_p}|\langle f,\psi_a\rangle| \le (\sum_{a \in \mathbb{Q}_p/\mathbb{Z}_p}|\langle T(f),\psi_a\rangle|^2)^{1/2}(\sum_{a \in \mathbb{Q}_p/\mathbb{Z}_p}\frac{1}{|c(a)|^2})^{1/2} = C \|T(f)\|_2$$ so $(1)$ would converge absolutely.
$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}$The Fourier series need not converge absolutely. First, some notation. Let $A_n = p^n \ZZ_p \subseteq \ZZ_p$ and let $B_n = p^{-n} \ZZ_p/\ZZ_p \subset \QQ_p/\ZZ_p$. Let $\chi_S$ denote the characteristic function of a set $S$, meaning that $\chi_S(s)=1$ if $s \in S$ and $\chi_S(s)=0$ otherwise. Then the Fourier transform of $\chi_{A_n}$ is $p^{-n} \chi_{B_n}$.
Let $r_n$ be any sequence of real numbers such that $\sum_{n=0}^{\infty} r_n$ is conditionally convergent. Then $\sum_{n=0}^{\infty} r_n \chi_{A_n}$ is a continuous function on $\ZZ_p$. (Because, if $s \in A_m \setminus A_{m+1}$, then $\sum_{n=0}^{\infty} r_n \chi_{A_n}(s)=\sum_{n=0}^m r_n$. If $s \to 0$ then $m \to \infty$, so this approaches $\lim_{m \to \infty} \sum_{n=0}^m r_n$, which is the definition of $\sum_{n=0}^{\infty} r_n$.)
The Fourier transform of $\sum_{n=0}^{\infty} r_n \chi_{A_n}$ is $\sum_{n=0}^{\infty} r_n p^{-n} \chi_{B_n}$. If $t \in B_m \setminus B_{m-1}$, then this is $\sum_{n=m}^{\infty} r_n p^{-n}$ (which is absolutely convergent, since $r_n \to 0$). If we sum over all points $t \in B_m \setminus B_{m-1}$, we get $$c_m:=(1-1/p) \sum_{n=m}^{\infty} r_n p^{m-n}.$$ If the Fourier series were absolutely convergent, then in particular we could clump together the $t$'s in the same $B_m \setminus B_{m-1}$ and then order the $m$'s arbitrarily. In other words, $\sum c_m$ would be aboslutely convergent.
But it is easy to come up with $r_n$ such that $\sum c_m$ is not absolutely convergent. For example, the standard $\tfrac{(-1)^n}{n}$ works.
Of course, what you wanted was a condition under which the sum would be absolutely convergent, but you seemed unclear as to whether this would always happen, so it seemed worth pointing out that the answer is "no". I suspect there should be some sort of condition in terms of the modulus of continuity of the function; I'll think about it a bit.