Yes. Let the random variables $V_i$ be independent, with $V_i$ uniform on $[-\frac 1i,\frac 1i]$. Let $N$ be an independent standard normal. Inductively define $Z_i\in \{\pm 1\}$ by
$$
Z_i=
\begin{cases}
1&\text{if $\text{sgn} (N-\sum_{j<i}Z_jV_j)=\text{sgn}(V_i)$}\\
-1&\text{otherwise.}
\end{cases}
$$
Notice that $\sum Z_iV_i=N$ almost surely - since the $|V_i|$ approach 0, but are almost surely not summable, the partial sums overshoot successively in one direction then the other, converging to $N$.

An observation that we need is the following: the involution that negates all of the $V$’s and $N$ preserves probabilities; and the values of the $Z$ variables are unchanged. In particular for any $a$ and $i$, we have $\mathbb P(V_i<a,\, Z_i=-1)=\mathbb P(V_i>-a,\, Z_i=-1)$.

Now we claim that $U_i=Z_iV_i$ has the same distribution as $V_i$.
To see this, notice that $\mathbb P(U_i<a)=\mathbb P(V_i<a,\,Z_i=1) +
\mathbb P(V_i>-a,\, Z_i=-1)$. But by the above observation, $\mathbb P(V_i>-a,\,Z_i=-1)=\mathbb P(V_i<a,\,Z_i=-1)$, so that $\mathbb P(U_i<a)=\mathbb P(V_i<a)$ as required.

**added comment**: @Michael Hardy pointed out the use of conditional convergence in this answer. It turns out this is essential: there does not exist a collection of dependent uniform random variables where the sum is (non-trivially) normal and the convergence is absolute. To see this, note that if $\sum a_i<\infty$, then the sum is compactly supported, so cannot be normal. If $\sum a_i=\infty$, then $\mathbb E\sum_{i=1}^M|U_i|=\sum_{i=1}^M a_i/2$ by linearity of expectation. Since $\sum_{i =1}^M |U_i|\le \sum_{i=1}^M a_i$, we deduce $\mathbb P(\sum_{i=1}^M |U_i|\ge \sum_{i=1}^M a_i/4)\ge 1/4$. In particular, the limit superior of these sets has measure at least $1/4$, but on this set the sum of the absolute values diverges.