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.
Best Answer
I'm pretty sure it's $0$ too.
One way to see this: given $2$ is a primitive root,
the quadratic residues (besides $0$) are $2^2, 2^4, 2^6, \dots$, and $2^{100}$,
so their sum is $2^2+2^4+2^6+\cdots+2^{100}=\dfrac{2^{102}-2^2}{2^2-1}=\dfrac23(2^{101}-2),$
which is a multiple of $101$ by Fermat's little theorem.
Another way to see this: since $101\equiv1\pmod4,$ $-1$ is a quadratic residue mod $101$,
so for every quadratic residue in the sum,
its additive inverse $\bmod 101$ is also a quadratic residue in the sum,
so the sum is $0$.