On the determinant $\det[(\frac{i^2+dj^2}p)]_{0\le i,j\le(p-1)/2}$ with $(\frac dp)=-1$ Let $p$ be an odd prime. For $d\in\mathbb Z$ we define
$$T(d,p):=\det\left[\left(\frac{i^2+dj^2}p\right)\right]_{0\le i,j\le(p-1)/2},$$
where $(\frac{\cdot}p)$ is the Legendre symbol. 
By (1.17) of my paper arXiv:1308.2900, if$(\frac dp)=-1$ then $(\frac{T(d,p)}p)=1$. 
Suppose that $p\equiv3\pmod4$. Then, by (1.14) of arXiv:1308.2900,  $T(d,p)=T(-1,p)$ for any $d\in\mathbb Z$ with $(\frac dp)=-1$. As $T(-1,p)$ is a skew-symmetric determinant of even order, it is an integer square.
In the case $p\equiv1\pmod4$, if $d$ and $d'$ are both quadratic nonresidues modulo $p$, then we clearly have $T(d,p)=\pm T(d',p)$.
I have the following conjecture which seems quite challenging.
Conjecture. Let $p\equiv1\pmod4$ be a prime and write $p=x^2+4y^2$ with $x$ and $y$ positive integers. Then, for any integer $d\in\mathbb Z$ with $(\frac dp)=-1$, there is a positive integer $t(p)$ (not depending on $d$) such that $$|T(d,p)|=2^{(p-1)/2}t(p)^2y.$$
Via Mathematica, I find that
\begin{gather}t(5)=1,\ t(13)=3,\ t(17)=4,\ t(29)=91,\ t(37)=81,\ t(41)=180,
\\t(53)=1703,\ t(61)=87120,\ t(73)=16104096,\ t(89)=3947892146,
\\ t(97)=19299520512,\ t(101)=885623936875,\ t(109)=36548185365.\end{gather}
Your comments are welcome!
PS: I have verified the conjecture for all primes $p<5000$ with $p\equiv1\pmod4$.
 A: I have obtained some partial results that seem promising, but not a full solution:
Let $\chi$ be a nontrivial Dirichlet character mod $p$ with $\chi(-1)=1$. Then
$\sum_{j=0}^{(p-1)/2} \left( \frac{ i^2+ dj^2}{p} \right) \chi(j) = \frac{1}{2} \sum_{j=1}^{p-1 } \left( \frac{ i^2+ dj^2}{p} \right) \chi(j) $
If $i=0$, then this sum is zero. Otherwise, we can perform a substitution replacing $j$ by $ij$, getting 
$ \frac{1}{2} \sum_{j=1}^{p-1 } \left( \frac{ i^2+ di^2 j^2}{p} \right) \chi(i j) =  \chi(i) \frac{1}{2}\sum_{j=1}^{p-1 } \left( \frac{ 1+ d j^2}{p} \right) \chi( j)  $. So $\chi$ is an eigenvector  of this matrix, with eigenvalue $ \frac{1}{2}\sum_{j=1}^{p-1 } \left( \frac{ 1+ d j^2}{p} \right) \chi( j)  $. 
The complement to these eigenvectors has a basis consisting of the function that is $1$ on $0$ and the function that is $1$ on all nonzero $j$. The matrix preserves the space generated by this basis, and acts on it by $$\begin{pmatrix} 0 & - (p-1)/2 \\ 1 & \frac{1}{2} \sum_{j=1}^{p-1}  \left( \frac{1+ dj^2}{p} \right)\end{pmatrix} $$ and therefore with determinant $(p-1)/2$. So the determinant of your matrix is
$$\frac{p-1}{2} \prod_{\chi \textrm { nontrivial, } \chi(-1)=1} \left( \frac{1}{2}\sum_{j=1}^{p-1 } \left( \frac{ 1+ d j^2}{p} \right) \chi( j) \right)$$
We can write $\sum_{j^2 = t} \chi(j)$ as $\chi_1(t) + \chi_2(t)$ where $\chi_1$ and $\chi_2$ are the squareroots of $t$ in the ring of characters, so $$\sum_{j=1}^{p-1 } \left( \frac{ 1+ d j^2}{p} \right) \chi( j)  = \sum_{t=1}^{p-1} \left( \frac{ 1+ d t}{p} \right) \chi_1( t)+ \sum_{t=1}^{p-1} \left( \frac{ 1+ d t}{p} \right) \chi_2( t) $$
If we focus attention on the eigenvector associated to $\chi$ the Legendre symbol, then $\chi_1$ and $\chi_2$ are the two characters of order $4$, so the left term is of the form $a+bi$ and the right term is its complex conjugate $a-bi$. By the evaluation of the absolute value of Jacobi sums, $a^2+b^2=p$, and because the number of $t$ with $\chi_1(t)= \pm 1$ (i.e. $t$ a quadratic residue) and $1+dt \neq 0$ (implied by $t$ a quadratic residue) is $p-1/2$, which is even, $a$ is even, so in fact $a=\pm y,b=\pm x$, and this eigenvalue is $\pm y$.
So it seems when the eigenvalue associated to the Legendre symbol is removed, you would like to have a square determinant. But I don't yet quite see how to obtain that.
I tried to get a square by finding matching pairs of eigenvalues, or by removing this eigenvector and conjugating to a skew-symmetric matrix, but neither approach quite worked.
