Let $U_a$ and $U_b$ be independent random variables with uniform distribution over $(0,1)$. Then $Y:=U_a-U_b$ has triangular distribution with density
$$ f_Y(x) = (1+x)\mathsf 1_{(-1,0)}(x) + (1-x)\mathsf 1_{(0,1)}(x). $$
Now let $Y_i$, $i=1,2,\ldots$ be i.i.d. with the above distribution. What is the distribution of $X_n:= \sum_{i=1}^n Y_i$?
I doubt there is a nice closed form for the density of $X_n$. Even for $n=2$, there are four cases to consider for integral bounds for the convolution, and the result ends up being $$ f_{X_2}(t) = \frac16(8+12t+6t^2+t^3)\mathsf 1_{(-2,1)}(t) + \frac16(4-6t^2-3t^3)\mathsf 1_{(-1,0)}(t) + \frac16(4-6t^2+3t^3)\mathsf 1_{(0,1)}(t) + \frac16(8-12t+6t^2-t^3)\mathsf 1_{(1,2)}(t). $$ So in general, the density of $X_n$ is going to be a piecewise polynomial of degree $n+1$ over $(-n,-n+1),\ldots,(n-1,n)$. We can, however, find the characteristic function of $X_n$. First we compute \begin{align} \varphi_Y(\theta) :&= \mathbb E[e^{i\theta Y}]\\ &= \int_{-1}^0 e^{i\theta t}(1+t)\ \mathsf dt + \int_0^1 e^{i\theta t}(1-t)\ \mathsf dt\\ &= -\frac{e^{-i \theta } \left(1-e^{i \theta }\right)^2}{\theta ^2}. \end{align} Since the characteristic function of the sum of independent random variables is the product of their characteristic functions, it follows that \begin{align} \varphi_{X_n}(\theta) :&= \mathbb E[e^{i\theta X_n}]\\ &= \mathbb E[e^{i\theta\sum_{i=1}^n Y_i}]\\ &= \prod_{i=1}^n \mathbb E[e^{i\theta Y_i}]\\ &= (\mathbb E[e^{i\theta Y}])^n\\ &= \left(-\frac{e^{-i \theta } \left(1-e^{i \theta }\right)^2}{\theta ^2}\right)^n. \end{align}