The usual proofs are either constructive or can be made constructive fairly easily, sometimes by a slight weakening of the theorem. For example, let us read through this note by Brent Nelson. (Please read the four page proof before reading the rest of the answer.)
The first part, establishing that the Weierstraß function is uniformly continuous, is constructive.
The second part proves non-differentiability by demonstrating that the limit $\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$ does not exist. It is not an appeal proof by contradiction but a constructively valid refutation by contradiction.
There is just one sticky point, namely the proof tells us to select, for each $m \in \mathbb{N}$, an integer $\alpha_m$ such that $$-1/2 < b^m x_0 - \alpha_m \leq 1/2.$$
These bounds are too tight, because constructively we cannot show that every real is within distance $1/2$ of an integer. (When $b^m x_0$ is very close to $1/2$ we have to be able to decide whether it is actually smaller than $1/2$ or not in order to determine whether $\alpha_m = 0$ or $\alpha_m = 1$.)
So let us relax the condition for $\alpha_m$ a bit and see if the rest of the proof goes through. Let $q > 0$ be a small "relaxation" rational (we shall impose additional conditions on it later as needed). Using countable choice, we can select integers $\alpha_m$ such that
$$-1/2 - q < b^m x_0 - \alpha_m < 1/2 + q.$$
This is possible because $q$ gives us a bit of wiggle room.
Now the proof proceeds by showing that two partial sums, called $S_1$ and $S_2$, are respectively very small and very large. Smallness of $S_1$ is unproblematic. The proof that $S_2$ is large depends on bounding $x_m = b^m x_0 - \alpha_m$ away from $1$ so that
$$
\frac{1 + \cos(b^n \pi x_m)}{1 + x_m}
$$
is seen to be positive. This is still the case with our $q$ thrown in as long as $q < 1/2$. But where the original proof gets the bound $2/3$ we obtain a slightly worse estimate
$$
\frac{1 + \cos(b^n \pi x_m)}{1 + x_m} \geq
\frac{1}{1 + \frac{1}{2} + q} = \frac{2}{3 + q}.
$$
We can still salvage the rest of the proof by picking a small enough $q$: so long as $a b > 1 + (3 + q) \pi / 2$, the estimates in the rest of the proof will work. Since we assumed $a b > 1 + 3 \pi / 2$, there is a rational $q > 0$ such that $a b > 1 + (3 + q) \pi / 2$, so we use one such. (This was not an application of choice, because "choosing one thing known to exist" is formally an elimination of $\exists$.)
One would likely have to work a bit harder to also eliminate countable choice.
