Let $A_q=\{x^2:x\in\mathbb{F}_q^{*}\}$.
Let $\pi^{\prime}$ be the permutation on $A_q$ defined by $$\pi^{\prime}(a_k)=a_{\pi(k)}.$$
Then
$$
\sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}=\sum_{a\in A_q}(a\pi^{\prime}(a))
$$
so that if $\pi$ is the identity permutation
$$
\sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}=\sum_{a\in A_q}(a^2).
$$
Hence if $\pi$ is the identity permutation and $q$ is a prime congruent to $3\bmod 4$,
$$
\sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}=\sum_{a\in A_q}(a)=0.
$$
Let $a\not=b$ and $\pi^{\prime}_{a,b}$ be the transposition $\pi^{\prime}_{a,b}(a)=b$,$\pi^{\prime}_{a,b}(b)=a$ and $\pi^{\prime}_{a,b}(c)=c$, $c\not=a,b$.
Hence if $q$ is a prime congruent to $3\bmod 4$,
$$\sum_{m\in A_q}m\pi^{\prime}_{a,b}(m)=-(a+b)^2$$
where $a,b\in A_q$. Let $\pi^{\prime}_{a,b,c,d}$ be the product of two transpositions $\pi^{\prime}_{a,b}$, $\pi^{\prime}_{c,d}$, $a,b,c,d$ all distinct in $A_q$. Then if $q$ is a prime congruent to $3\bmod 4$,
$$
\sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}=\sum_{m\in A_q}(m\pi^{\prime}_{a,b,c,d}(m))=-(a+b)^2-(c+d)^2.
$$
So for $q$ a prime congruent to $3\bmod 4$,
$$
\{\sum_{k=1}^{(q-1)/2}a_ka_{\pi(k)}:\pi\in S_{(q-1)/2}\}=\mathbb{F}_q
$$
if every $k\in\mathbb{F}_q$ can be represented as 
$$
k=(a+b)^2+(c+d)^2
$$
where $a,b,c,d$ are distinct elements in $A_q$.