Consider the polynomial $$ P_n(t)=(-1)^n (2t-1) \frac{(2t-1)^{n-1}-t^{2n-2}}{(t-1)^2},$$ that is up to a factor $(-t^2)^{n-1}$ the Tutte polynomial $\frac{x^n-x}{x-1}$ of the cyclic graph evaluated in $\frac{2t-1}{t^2}$.
I claim that the coefficients of the polynomial $P_n(t)$ are $$ p_k = \begin{cases} (-1)^k \sum_{i=0}^k \binom{n-2-i}{k-i}(i+1)2^{k-i} & k\leq n-2\\ 2n-1-k & n-1\leq k \leq 2n-3 \end{cases} $$ In other words I need to prove the following equality $$P_n(t)= \sum_{k=0}^{2n-3} \left((-1)^k \sum_{i=0}^k \binom{n-2-i}{k-i}(i+1)2^{k-i} \right) t^k ,$$ where the binomial coefficient $\binom{-a}{b}$ is $(-1)^b \binom{a+b+1}{b}$ for all positive integers $a,b$.
A partial answer. By writing $$\begin{eqnarray*} P_n(t)&=&(-1)^{n+1}(2t-1)\frac{(t^2)^{n-1}-(2t-1)^{n-1}}{(t-1)^2}\\&=&(-1)^{n+1}\sum_{h=0}^{n-2}(2t-1)^{n-1-h} t^{2h}\tag{1} \end{eqnarray*}$$ we have $$ [t^k]P_n(t) = (-1)^{n+1}\sum_{h=0}^{n-2}[t^{k-2h}](2t-1)^{n-1-k}=\sum_{h=0}^{n-2}\binom{n-1-k}{k-2h}2^{k-2h} \tag{2}$$ where the RHS of $(2)$ is pretty close to the expression we want to get.
In the opposite direction, for a fixed value of $k\leq n-1$ $$ \sum_{i=0}^{k}\binom{n-2-k+(k-i)}{k-i}2^{k-i}(i+1)=[u^k]\left(\sum_{b\geq 0}(b+1)u^b\right)\cdot\left(\sum_{a\geq 0}\binom{n-2-k+a}{a}2^a u^a\right) $$ hence the LHS is the coefficient of $u^k$ in $\frac{1}{(1-u)^2(1-2u)^{n-k-1}}$ and $$ \sum_{k=0}^{n-1}u^k\sum_{i=0}^{k}\binom{n-2-i}{k-i}2^{k-i}(i+1) = \frac{1-(1-2u)^n}{2u(1-u)^2(1-2u)^{n-1}}\tag{3}$$ where the RHS of $(3)$ matches $P_n(u)$ up to the monomial $u^{n-1}$.
Now it should be enough to extend this approach to $k\geq n$ to prove the wanted identity.