When $p=3 \pmod 4$, there is still a simple formula for adding all of the primitive roots.
There probably is no easy answer here. The primitive roots of 7 are 3 and 5, adds to 8, and for 11, one has 2, 6, 7 and 8, which adds to 23.
One can establish that the primitive roots add up to a number modulo $p$, specifically $(-1)^m$, where $m$ is the number of distinct prime divisors of $p-1$, and if $p-1$ is divisible by a power of a prime, then the sum of primitive roots is a multiple of $p$.
The proof of this is fairly straight forward. Consider for example, $30$. The sum of the complete solutions of $x^n \pmod p$ is $0$, where $n \mid p-1$
So one works through the divisors. For $n=1$, the sum is 1. For $n=p_1$, q prime the sum is -1. For $n=p_1 p_2$, the sum is +1. The sum of all of the divisors of $p_1 p_2$, is then $f(1)+f(p_1)+f(p_2)+f(p_1 p_2) = 0$.
For powers of $p_n$, the sum $f(1)+f(p_1)+f(p^2_1) \dots$, is zero, so every term after the second must be zero.
Since the primitive roots is the largest divisor of $p-1$, then it is by that formula.
The cases where $p\in\{p_1,p_2,p_3\}$ are easy, and I assume that $p\ne p_i$, $i\in\{1,2,3\}$.
For brevity I write $P:=p_2p_3$; also, let $\left(\frac{\cdot}p\right)$ denote the Legendgre symbol.
If either $P$ or $-P$ is a quadratic residue mod $p$, then we can take $y=0$ and choose $x$ appropriately; suppose thus that both $P$ and $-P$ are quadratic non-residues. Notice that this implies $\left(\frac{-1}p\right)=1$.
If $p_1$ is a quadratic residue mod $p$, then we can take $x=0$; suppose therefore that $p_1$ is a quadratic non-residue mod $p$.
Clearly, with all the assumption made, it suffices to show that there exists $x\in\mathbb Z/p\mathbb Z$ such that one of $x^2-P$ and $x^2+P$ is a quadratic residue, while another is a quadratic non-residue mod $p$.
Suppose for a contradiction that for each $x\in\mathbb Z/p\mathbb Z$ we have $\left(\frac{x^2-P}p\right)=\left(\frac{x^2+P}p\right)$. Replacing $x$ with $Px$ and using multiplicativity of the Legendre symbol, we conclude that $\left(\frac{Px^2-1}p\right)=\left(\frac{Px^2+1}p\right)$, for all $x\in\mathbb Z/p\mathbb Z$. Recalling that $P$ is a quadratic non-residue, we further conclude that $\left(\frac{z+1}p\right)=\left(\frac{z-1}p\right)$ for any quadratic non-residue $z$, and also for $z=0$.
Let $N_0\subset\mathbb Z/p\mathbb Z$ be the set containing all quadratic non-residues and $0$. Since $|N_0|=(p+1)/2>p/2$, this set contains two consecutive elements of $\mathbb Z/p\mathbb Z$; say, $z-1$ and $z$. But then also $z+1\in N_0$ and, consecutively, $z+2\in N_0$ etc. As a result, all elements of $\mathbb Z/p\mathbb Z$ are in $N_0$, meaning that there are no quadratic residues mod $p$. This is a clear nonsense, showing that $\left(\frac{x^2-P}p\right)=\left(\frac{x^2+P}p\right)$ cannot hold for all $x\in\mathbb Z/p\mathbb Z$.
Notice that we have not used the assumption $p_1\equiv p_2\equiv p_3\equiv 5\pmod 8$ and $\left(\frac{p_i}{p_j}\right)=1$, $i\ne j$.
Best Answer
Easy way: $ $ applying my denesting formula we can compute $\color{#0a0}{\sqrt x}\,$ by completing a square
$\qquad \qquad\ 34x = x^2+1 \iff 36x = (x\!+\!1)^2\!\!\iff x = \left(\!\color{#0a0}{\dfrac{x\!+\!1}6}\!\right)^{\!2}$
Remark $ $ So $\:\!x\:\!$ has a $\:\!\rm\color{#0a0}{square\ root}\:\!$ in any ring where $\color{#0a0}{1/6}\,$ exists, e.g. $\:\!\Bbb Z_n\:\!$ for all $\:\!n\:\!$ coprime to $\:\!6,\,$ e.g. OP's $\,n=$ prime $p\equiv 7\pmod{\!8},\,$ where $\,p\neq 2,3\Rightarrow p\,$ coprime to $\:\!6$.