If
$$ h(k) = \begin{cases} \frac{1}{2l+1} & -l \leq k \leq l \\0 & \text{otherwise}\end{cases} $$
Where $l \geq 0$ is some integer. I've done some computation and the summation
$$ F[h](\omega)=\sum_{k=-\infty}^{+\infty} h(k)e^{j\omega k} = \frac{1}{2l+1}\frac{\sin(\omega(l+1/2))}{\sin(\omega/2)} = H(\omega) $$
If I consider $h$ convolved with itself $n$ times I have $$ F[h*h\ldots*h](\omega) = H^{n}(\omega) = \frac{1}{\left(2l+1\right)^n}\left(\frac{\sin(\omega(l+1/2))}{\sin(\omega/2)}\right)^n $$
Now I'd like to compute (in closed form) the integral
$$ I_n(k) = \frac{1}{2\pi}\int_{-\pi}^{\pi} H^n(\omega)e^{j\omega k} d\omega $$
The clue I had was using the residual theorem somehow, If I wanted to compute it exactly. However I have some difficulties with the calculation because of the dependencies with $n$. I would be happy with an asymptotic estimation as well since in practice I'm interested in large $n$. So the difficulty is given by finding an asymptotic approximation of
$$ \frac{1}{\left(2l+1\right)^n}\left(\frac{\sin(\omega(l+1/2))}{\sin(\omega/2)}\right)^n \approx \;? $$
(I guess when $n$ tends to infinity $I_n$ should look like a Gaussian).
Update: From this article in wikipedia I know that:
$$ 1 + 2\sum_{k=1}^l \cos(k\omega) = \frac{\sin(\omega(l+1/2))}{\sin(\omega/2)} $$
Which implies $$ \left( \frac{\sin(\omega(l+1/2))}{\sin(\omega/2)} \right)^n = \left( 1 + 2\sum_{k=1}^l \cos(k\omega) \right)^n $$
and I'm stuck again...
Update 2:
Not sure if this might lead to something. For $\omega\to 0$ we have
$$ f_l(\omega) = \frac{\sin(\omega(l+1/2))}{\sin(\omega/2)} = 1 - \frac{1}{6}(l+1)l\omega^2 + O(\omega^2) $$
And therefore I have
$$ \left( \frac{\sin(\omega(l+1/2))}{\sin(\omega/2)} \right)^n \approx \left(1 - \frac{1}{6}(l+1)l\omega^2 \right)^n \approx e^{-\left(\sqrt{\frac{1}{6}n(l+1)l}\omega \right)^2} $$
And therefore the integral I want is given by
$$ I_n(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} H^n(\omega)e^{j\omega k} d\omega \approx \frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{-\left(\sqrt{\frac{1}{6}n(l+1)l}\omega \right)^2} e^{j\omega k} d\omega = \int_{-\infty}^{+\infty} e^{-\left(\sqrt{\frac{1}{6}n(l+1)l}\omega \right)^2} e^{j\omega k} d \left( \frac{\omega}{2\pi} \right) $$
And defining $f = \frac{\omega}{2\pi}$, we end up with a known integral, such as this. I don't like much my argument, is really intuitive, but I believe it can be justified formally.
Update 3:
A last attempt and I'll give up for now.
Trying to expand
$$ g_l(\omega)^n = \left( \frac{1}{2l+1}\frac{\sin(\omega(l + 1/2))}{\sin(\omega/2)} \right)^n $$
Let me start by defining
$$ g_l(\omega) = \frac{1}{2l+1}\frac{\sin(\omega(l + 1/2))}{\sin(\omega/2)} $$
using the equality mentioned in the update 1 I can write
$$ \left\{ \begin{array}{l} g_l(0) = 1 \\ g^{(1)}_l(0) = 0 \\ g^{(2)}_l(0) = -\frac{(l+1)l}{3} \\ \end{array} \right. . $$
For $g_l(\omega)^n$ we have
$$ \left\{ \begin{array}{l} \left( g^n_l(\omega)^n \right)^{(1)} = n g^{n-1}_l(\omega) g^{(1)}(\omega) \\ \left( g^n_l(\omega)^n \right)^{(2)} = n g^{n-2}_l(\omega) \left( (n-1)g^{(1)}(\omega) + g^{(2)}(\omega) \right) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} g^n_l(0) = 1 \\ \left( g^n_l(0)^n \right)^{(1)} = 0 \\ \left( g^n_l(0)^n \right)^{(2)} = -\frac{n(l+1)l}{3} \end{array} \right. $$
And we have
$$ g_l(\omega)^n = 1 - \frac{n(l+1)l}{6} \omega^2 + O(\omega^2) $$
There's a different way to look at the problem that involves the Fourier transform implicitly. Assuming $h(k)$ models a probability distribution, consider the sequence of $n$ random variables $X_1,\ldots,X_n$ distributed according $h$. Easily we can find out that
$$ \overline{X_i} = 0, Var(X_i) =\frac{(l+1)l}{3}. $$
Consider now
$$ S_n = \frac{X_1 + \ldots + X_n}{n} $$
By the CLT we now $S_n$ will be have distribution
$$ S_n \sim \frac{1}{\sqrt{2 \pi \frac{(l+1)l}{3n}}} e^{-\frac{x^2}{2\frac{(l+1)l}{3n}}} = \mathcal{N}\left(0,\frac{(l+1)l}{3n}\right) $$
Defining $Q_n = n S_n$ by a known theorem in probability theory we can assert
$$ Q_n \sim \mathcal{N}\left(0,\frac{n(l+1)l}{3}\right) $$
Therefore my original problem translates into finding the Fourier transform of the distribution of $Q_n$, which can be easily found in this case using the link in update 1.