A general useful fact (compare with my answer to your previous question) is that whenever we have $A=\{a_1,\dots,a_n\}\subset \{0,1,\dots,p-1\}$ such that $n=|A|$ is odd, the sign of a product $\prod_{i<j}(\{a_j-\theta\}_p-\{a_i-\theta\}_p)$ does not depend on the choice of remainder $\theta$ modulo $p$. This may be proved using the observation 
$$
{\rm sign}\, \left(\{b-\theta\}_p-\{a-\theta\}_p\right)={\rm sign}\, \left(\{b\}_p-\{a\}_p\right)\cdot (-1)^{\chi(\{b\}_p<\theta)+\chi(\{a\}_p<\theta)}.
$$
When you multiply this by all pairs $b=a_j,a=a_i,i<j$, each multiple $(-1)^{\chi(\{a\}_p<\theta)}$ appears exactly $n-1$ times, which is even.

Now you take $n=(p-1)/2$, $a_j=T_{(p-1)/2-j}=\{-1/8+j^2/2\}_p$, $j$ varies from 0 to $(p-3)/2=n-1$ and you look for the sign of $\prod_{0\leqslant i<j\leqslant n-1} (a_i-a_j)$ (the order is inversed). This is the same as $$(-1)^{n\choose 2}\cdot {\rm sign}\, \prod_{0\leqslant i<j\leqslant n-1} (a_j-a_i)=(-1)^{n\choose 2}\cdot {\rm sign}\, \prod_{0\leqslant i<j\leqslant n-1} \left(\{j^2/2\}_p-\{i^2/2\}_p\right).$$
It is convenient to add $j=n$ and consider the product
$$
\prod_{0\leqslant i<j\leqslant n} \left(\{j^2/2\}_p-\{i^2/2\}_p\right).
$$
For finding its sign we exclude $i=0$ which does not rely on the sign and consider two cases. 

1) $p$ is congruent to 7 modulo 8. In this case $2$ is a quadratic residue and the map $x\mapsto x/2$ permutes the (nonzero) quadratic residues. This permutation is even, because all cycles have the same odd length (dividing odd number $(p-1)/2$). On the other hand the sign of this permutation equals 
$$
{\rm sign}\,\prod_{1\leqslant i<j\leqslant n} \frac{\{j^2/2\}_p-\{i^2/2\}_p}{\{j^2\}_p-\{i^2\}_p}.
$$
Therefore the numerator and the denominator have the same sign and we reduced the problem to the already solved in your paper.

2) $p=8k+3$. In this case -2 is a quadratic residue and we similarly get
$$
{\rm sign}\,\prod_{1\leqslant i<j\leqslant n} \frac{\{-j^2/2\}_p-\{-i^2/2\}_p}{\{j^2\}_p-\{i^2\}_p}=1.
$$
It remains to note that ${\rm sign}\, (\{j^2/2\}_p-\{i^2/2\}_p)=-{\rm sign}\, (\{-j^2/2\}_p-\{-i^2/2\}_p)$ (and mind the multiple $(-1)^{n\choose 2}$, but it equals 1 for $p=8k+3$, $n=4k+1$).

The last thing to do is to study what we have added: the sign of $\prod_{0\leqslant i<n}(\{n^2/2\}_p-\{i^2/2\}_p)$. We have $n^2/2\equiv 1/8$. Again consider two cases.

1) $p=8k+7$, then $1/8\equiv (p+1)/8$ and we look for the number of quadratic residues (recall that $i^2/2$ is a quadratic resdiue) greater than $(p+1)/8$. This has the parity different from that of the number of quadratic residues at most $(p+1)/8$ (since the total number of quadratic residues is odd.)  But we had also a sign $(-1)^{n\choose 2}=-1$ before. So we get your conjecture in this case.

1) $p=8k+3$. Then $1/8\equiv (5p+1)/8$ and we look for the number of quadratic non-residues greater than $(5p+1)/8$. This is the same as the number of quadratic residues less than $(3p-1)/8$.