Does absolute convergence of a sum imply uniform convergence?

There are easy counterexamples to the statement, which others have already supplied.

But let me say what mistake I think you're making. Dini's Theorem has a lot of fine print: in particular, it requires the pointwise limit function to be continuous. You can see that this hypothesis is not met in the above counterexamples. Conversely, if each $f_n$ and also $\sum_{n=0}^{\infty} |f_n(x)|$ were continuous, then indeed you could apply Dini's Theorem to see that the convergence of $\sum_n |f_n(x)|$ is uniform, from which it follows that the convergence of $\sum_n f_n(x)$ is uniform, say via the Cauchy criterion.

You have my sympathies for forgetting this hypothesis of Dini's Test. I still have a set of lecture notes up on my webpage where I state Dini's Test incorrectly (I can't find the tex file to change it!), and when I mentioned it in a second course more recently, I made my way through the statement and the proof (!) without making explicit the need for continuity of the limit function. Finally I got it right: see for instance $\S III.1.4$ of these notes.

The example given by Davide Giraudo works well. Indeed for $x \neq 1$ we have $$\sum_{n=0}^\infty x^n(1-x) = \sum_{n=0}^\infty x^n - \sum_{n=1}^\infty x^{n} = 1,$$ but for $x = 1$ the series is $0$. If the convergence was uniform, the resulting function would be continuous, which it isn't.

There is however a result called Weierstrass M-test. It says that if each of the functions $f_n$ is bounded, i.e. $\sup_{x \in [0,1]} |f_n(x)| \le M_n$ for some constant $M_n > 0$, and if $\sum_{n=0}^\infty M_n < \infty$, then the convergence is uniform. (Here I have taken $[0,1]$ as the domain of the functions. You may replace this by any set.)

In view of the reference to Dini one might ask, what if all $f_n$ and $f$ are continuous functions on a compact set $K$? Thus if all $f_n \ge 0$, Dini would say the convergence is uniform on $K$. But without that hypothesis it's not true. Consider $K=[0,1]$ where
$f_{2n-1}$ is the piecewise-linear interpolation of $f_{2n-1}(0) = 0$, $f_{2n-1}(1/(2n)) = 1$, $f_{2n-1}(1/n) = 0$, $f_{2n-1}(1) = 0$, while $f_{2n}(x) = - f_{2n-1}(x)$. Then $\sum_{n=1}^\infty f_n(x)$ converges absolutely to $f(x) = 0$ but the odd partial sums have maximum value 1.