The first part of the conjecture follows from a result of Robin Chapman. Letting $C_{p}(x)$ be the $\frac{p-1}{2}$ by $\frac{p-1}{2}$ matrix with $(i,j)$ entry $x+(\frac{i+j-1}{p})$, where $x$ is an indeterminate, he proved that when $p\equiv 3 \pmod 4$,
$$\det C_{p}(x)=-2^{(p-1)/2}x.$$
If we let $D_{p}(x)$ be the reflection of $C_{p}(x)$ w.r.t. the antidiagonal, then we have
$$\det C_{p}(x)=\det D_{p}(x)$$
and
$$D_{p}(x)=\left[x+\left(\frac{(p+1)/2-j+(p+1)/2-i-1}p\right)\right]_{1\le i,j\le(p-1)/2}=\left[x-\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}$$
Hence
$$\det \left[x-\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=-2^{(p-1)/2}x.$$
So
$$\det \left[x+\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=(-1)^{(p-1)/2}\det \left[-x-\left(\frac{i+j}p\right)\right]_{1\le i,j\le(p-1)/2}=(-1)(-2^{(p-1)/2}(-x))=-2^{(p-1)/2}x$$