Quick background
Ingham gives the density of zeros of $\zeta(s)$ as
$$N(T) =\frac{T}{2\pi}\log\left(\frac{T}{2\pi}\right)-\frac{T}{2\pi}+O(\log T)\hspace{10mm}(1) $$
meaning the complex part $\gamma_k$ of $s_k=\sigma_k + i\gamma_k$ does not exceed T.
Disregarding $O(\cdot)$, N(T) doesn't exceed $T$ until around $k=9100$ and if we use actual zeros $N(T)>T$ at $\gamma_{9141}.$
Question
The density estimate suggests that the cardinality of $\gamma_k$ greatly exceeds T for large T. For example, the 10 millionth zero ($\gamma)$ has a numerical value of about 5 million. And of course these $\gamma_k$ are more frequent in, say, $(T/2, T)$ than $(0,T)$ as T grows.
So intuitively I would expect that $\frac{1}{N(T)}\sum_{\gamma_k\leq T} \gamma_k\sim T$ or, what is about the same, suppressing the index, $\frac{1}{N(T)}\sum_{|s|<T}|s|\sim T.$
If the reasoning is sound then I would also expect that (even faster as T grows),
$$\frac{1}{N(T)}\sum_{|s|<T}\log |s|\sim \log T $$
or, putting $x$ for $T$ and $n(x)$ for $N(T)$,
$$\frac{1}{n(x)}\sum_{|s|<x}\log|s|\sim \log x $$
and via (1) above and the p.n.t. a connection between the zeros and $\pi(x),$
$$\pi(x)\sim \frac{x}{\log x} \sim \frac{x}{\frac{1}{n(x)}\sum_{|s|<x}\log|s|}\hspace{10mm}(2)$$
A numerical example. For $n(x)=10000, \sum_{\gamma< x}\log (\gamma)\approx 83545.3,x =\gamma_{10000}\approx 9877.8$.
So the r.h.s. of (2) above works out to about 1182.3, while $\pi(10000) = 1229.$ FWIW this is in the ballpark, since $n/\log n \approx 1085.$
The argument is basically that the mean of the zeros is shifting toward $x$ as $x\to \infty.$ Before trying to formalize this I wonder if the idea seems sound to this point?
I alluded to partial summation in the comments, but in fact we can do this even more easily by using the fact that $\log \gamma$ grows very slowly. It's clear that the average value of $\log \gamma$ for $0 < \gamma \le T$ is at most $\log T$, so to establish the asymptotic it is enough to show that
$$\sum_{\gamma\le T}\log |\gamma| > (1 - o(1))\, N(T) \log T.$$
To see this, just start the summation at some $f(T)$ where $f(T)$ is small enough that $f(T) = o(T)$ but large enough that $\log f(T) \sim \log T$: for instance, $f(T) = T/\log T$ will do. Then
$$\sum_{\gamma\le T}\log |\gamma| > \sum_{ f(T) < \gamma \le T}\log |\gamma| > [ N(T)-N(f(T)) ] \log f(T).$$
It is easy to check that $N(T/\log T) = o(N(T))$ and that $\log(T/\log T) \sim \log T$, so the first term is $\sim N(T)$ and the second term is $\sim \log T$. And we're done.
[Note that this only uses the main term of the estimate for $N(T)$, so the strong error bound is not even necessary for this result. log averages are very well-behaved.]