Here is a simple explicit example of a measure $\mu$ with all convolution powers singular: let $\mu$ be the distribution of
$$
\sum_{k=1}^\infty X_k 2^{-k!},
$$
where the $X_k$ are IID taking values $0$ and $1$ with equal probability. The support of $\mu$ is the set $A$ of points whose binary expansion has non-zero digits only at places $k!,k\in\mathbb{N}$. Using this, it follows easily that for each $n$, the $n$-fold arithmetic sum $A^{+n}=A+\ldots+A$ has zero Lebesgue measure (in fact it has zero Hausdorff dimension), since the binary expansion of points in $A^{+n}$ must have zeros at all places not of the form $k!+j$ for some $0\le j\le n$. Since the convolution power $\mu^{*n}$ is supported on $A^{+n}$, it must be purely singular. On the other hand, $\mu$ has no atoms so neither does $\mu^{*n}$. This example doesn't use the Fourier transform at all.

Let me also give complete details of why the Cantor-Lebesgue measure $\mu$ on the middle-thirds Cantor set is an example of a measure all of whose self-convolutions are purely singular (but continuous, since $\mu$ itself already has no atoms).

Recall that $\mu$ can be realized as the distribution of the random sum $\sum_{i=1}^\infty X_i 3^{-i}$, where $X_i$ are independent random variables taking the values $0$ and $2$ with equal probability. In particular,
$$
\widehat{\mu}(\xi) =\prod_{i=1}^\infty \cos(2\pi 3^{-i}\xi)
$$
from which it easily follows that $\widehat{\mu}(\xi)$ doesn't decay to $0$ as $\xi\to\infty$ (this also follows from the fact that $\mu$ is invariant under multiplication by $3$ on the circle). Hence, if $\mu^{*n}$ denotes the $n$-fold self-convolution, then $\widehat{\mu^{*n}}(\xi)=\widehat{\mu}(\xi)^n$ also doesn't decay to $0$, so $\mu^{*n}$ cannot be absolutely continuous by the Riemann-Lebesgue Lemma.

Now, recall that a Borel probability measure $\nu$ on the real line is called *self-similar* if there are contracting similarities $f_1,\ldots,f_m$ and a probability vector $(p_1,\ldots,p_m)$ such that
$$
\nu(A)= \sum_{j=1}^m p_i\, \nu(f_j^{-1}(A)).
$$
Given the tuples $(f_1,\ldots,f_m)$ and $(p_1,\ldots,p_m)$ there is exactly one Borel probability measure satisfying the above identity; this follows from seeing the right-hand side as a contracting operator on an appropriate Banach space (actually we only care about uniqueness, so we only need to have a contraction, no completeness is required).

If $\nu$ is a self-similar measure as above, then the absolutely continuous and singular parts of $\nu$ are seen to satisfy the same self-similarity relation. By the uniqueness of $\nu$, one of these parts has to be trivial. So self-similar measures are either absolutely continuous or purely singular with respect to Lebesgue.

To conclude, I claim that $\mu^{*n}$ is a self-similar measure for every $n$. Since we already established that $\mu^{*n}$ cannot be absolutely continuous, this will show that it must be purely singular.

By definition, $\mu^{*n}$ is the distribution of the random sum
$$
\sum_{i=1}^\infty Y_i^{(n)} 3^{-i},
$$
where the $Y_j^{(n)}$ are IID with the distribution of the sum of $n$ independent realizations of $X_i$ (defined above). In particular, the distribution of $Y_i^{(n)}$ has the form $\sum_{j=0}^{n} p_j^{(n)} \delta_{2j}$ for some probability vector $p_j$. It can then be easily checked that $\mu^{*n}$ is the self-similar measure corresponding to the contractions $(x/3+2j)_{j=0}^{n}$ an the probability vector $(p_j)_{j=0}^n$.

Let me mention that it follows from work of Garsia in the 60s that not only $\mu^{*n}$ is singular for all $n$, but in fact for each $n$ there is a Borel set $A_n$ of Hausdorff dimension strictly less than $1$ such that $\mu^{*n}(A_n)=1$ (the topological support of $\mu^{*n}$ is an interval for all $n\ge 2$). On the other hand, the Hausdorff dimension of $A_n$ must tend to $1$; this follows from recent work of M. Hochman.