John Derbyshire in his book PRIME OBSESSION says on page 343:
"I’ll round off with a complete calculation of $\pi(1,000,000)$, the number of primes up to one million, using Riemann’s formula -- not for the fun of it, though it is of course great fun, but to make some important points about the error term."
He finds that secondary terms contribute an error of -29.37378.
My question:
How many zeta zeros are needed to find it?
Monte-Carlo Approximation of the Prime Counting Function
This is a numerical study of the dependence of the error $\delta\pi(x)$ in the prime counting function when $N$ zeros $\rho$ of the zeta function are used in the sum $\sum_\rho {\rm Li}\,(x^\rho)$.
Figure 3 shows that for $x=10^{12}$ the error $\delta\pi(x)$ is about 1100 for $N=10^3$ and 300 for $N=10^5$. The error drops by roughly a factor of two when the number of zeros is increased by a factor of ten. More extensive numerical data, with $x$ in the range $10^{10}$ to $10^{17}$, indicates that to reach the precise value of $\pi(x)$ one needs to include about $x^{1.38}$ zeros.
The value $x=10^6$ in the OP is well below this range, but as a first estimate I would conclude that to obtain $\pi(10^6)=78498$ accurate to the last of its five digits one would need $N\simeq 10^8$ zeros.
This is a rough estimate , about 101157 zeros . Using Zeta zeros and not zeros that are calculated by XI function or functional equation !
