The formula for $p_n$ was derived to be a probability relating two independent transformed binomials. In particular, $$p_n = \mathbb{P}(|U_n|<|V_n|)$$ where $$ U_n = a_1 + \dots a_k \hspace{20pt} V_n = b_1 + \dots + b_l $$ and $k = \lfloor (n+1)/2 \rfloor, l = \lfloor n/2 \rfloor$.
When $n=2m$ then $k = l = m$ and so $U_{2m}$ and $V_{2m}$ are i.i.d. In this case we can exploit symmetry; $\mathbb{P}(|U_n| < |V_n|) = \mathbb{P}(|V_n| < |U_n|)$, and so we can write: $$ p_{2m} = \frac{1}{2} \left ( 1 - \mathbb{P} ( |U_{2m}| = |V_{2m}| ) \right ).$$ In general, for discrete i.i.d. random variables $X, Y$ taking values in $\mathbb{N}$, we have: $$\mathbb{P}(X=Y) = \sum_{n = 0}^\infty \mathbb{P}(X = n)^2.$$ We can continue by splitting into two further cases depending on whether $m$ is even or odd, i.e. whether $n$ is $0$ or $2$ mod 4. The case $n = 4m + 2$ is easier; recall that $U_{4m+2} = 2B_{2m+1} - (2m+1)$ in distribution, where $B_{2m+1} \sim \text{Binomial}(2m+1,1/2)$. So \begin{align} \mathbb{P}(|U_{4m+2}| = |V_{4m+2}| ) & = \sum_{p=0}^m \mathbb{P}(B_{2m+1} = (m-p) \text{ or } (m+p+1) )^2 \\ &= 2^{-4m} \sum_{p=0}^m {2m+1 \choose p}^2\\ &= 2^{-4m-1} {4m + 2 \choose 2m+1}. \end{align} The case $n = 4m$ is similar, but the probability is slightly different since one has to take into account the event $\{|U_{4m}| = 0\}$. Then we have: \begin{align} \mathbb{P}(|U_{4m}| = |V_{4m}|) &= 2^{-4m} \left ( {2m \choose m}^2 + 4 \sum_{p=0}^{m-1} {2m \choose p}^2 \right ) \\ &=2^{-4m} \left ( 3 {2m \choose m}^2 + 2 {4n \choose 2n} \right). \end{align}