Let $W$ be the Weyl group of the root system of type $C_n$. Then $W$ can be identified with the group of signed permutations on $1, 2, ... , n$. Let $S = \{s_1, ... , s_n\}$, where $s_i$ swaps $i$ and $i+1$ for $1 \leq i \leq n -1$, and $s_n$ sends $n$ to $-n$.
Then $(W,S)$ is a Coxeter system. The long element $w_{\ell}$ of this system is the permutation which sends $i$ to $-i$ for all $i$. Is it possible to give a formula for a reduced decomposition of $w_{\ell}$ which works for all $n$?
Yes. By realizing $C_n$ as the root system of $\textrm{Sp}_{2n}$, one can calculate the length of $w_0$ as the number of positive roots. This is
$$\sum\limits_{i=1}^{n-1} i + \sum\limits_{i=1}^n i = \frac{(n-1)n}{2} + \frac{n(n+1)}{2} = n^2$$
A decomposition for the Weyl group element which switches $i$ and $n$ is
$$(i \space n) = s_i \cdots s_{n-2}s_{n-1}s_{n-2} \cdots s_i$$
The permutation which sends $i$ to $-i$ then has a decomposition
$$(i \space n)s_n(i \space n) = (s_i \cdots s_{n-1} \cdots s_i)s_n(s_i \cdots s_{n-1} \cdots s_i) = s_i \cdots s_{n-1}s_ns_{n-1} \cdots s_i$$
where we have used the fact that $s_i$ commutes with $s_n$ for $i = 1, ... , n-2$. Then our long element is the product (in any order)
$$w_{\ell} = \prod\limits_{i=1}^n s_i \cdots s_{n-1}s_ns_{n-1} \cdots s_i$$
The length of this decomposition is
$$\sum\limits_{i=1}^n [2(n-1 - (i-1)) + 1] = n + 2 \sum\limits_{i=1}^n(n-i) = n + 2n^2 - 2\frac{n(n+1)}{2} = n^2$$
which is the correct length.