I'll prove the general case in which $\mu$ is a positive measure on a space $X$ and $\mu(X) = 1$. Assuming that $\|f\|_q < \infty$ for at least one $0 < q < 1$, we have
$$
\lim_{p \to 0} \|f\|_{p} = \exp\left(\int_X \log|f| \,d\mu\right).
$$
Your particular case follows by setting $X = \{1, \ldots, n\}$ and $\mu(i) = 1/n$.
By definition
$$
\|f\|_p = \left\{\int_X |f|^p \,d\mu\right\}^{1/p}.
$$
Lemma 1: If $0 < r < s < 1$, then $\|f\|_r \le \|f\|_s$.
Proof: $\varphi(x) = x^{s/r}$ is a convex function. Apply Jensen's inequality to $\int_X |f|^r \,d\mu$ to get
$$
\left\{\int_X |f|^r \,d\mu\right\}^{s/r} \le \int_X |f|^s \,d\mu.
$$
Hence $\|f\|_r \le \|f\|_s$.
Lemma 2: If $0 < p < 1$, then $\int_X \log|f| \,d\mu \le \log \|f\|_p$.
Proof:
$\log$ is a concave function. Apply Jensen's inequality to $\int_X |f|^p \,d\mu$ to get the desired inequality.
From lemmas 1 and 2, it follows that $\log\|f\|_{1/n}$ is decreasing and bounded from below. Therefore, it converges as $n \to \infty$.
To find the limit, apply the inequality $\log a \le n(a^{1/n} - 1)$ with $a = \left\{\int_X |f|^{1/n}\,d\mu\right\}^{n} $ to get
$$
\log \|f\|_{1/n} \le \int_X \frac{|f|^{1/n} - 1}{1/n} \,d\mu. \tag{1}
$$
Use L'Hôpital's rule to obtain
$$
\lim_{x \to 0} \frac{a^x - 1}{x} = \log a.
$$
Take the limit of (1) as $n \to \infty$. Since $\dfrac{|f|^{1/n} - 1}{1/n}$ is dominated by the integrable function $\dfrac{|f|^{q} - 1}{q}$ for large enough $n$ and the value of $q$ stated in the assumptions, apply the dominated convergence theorem to get
$$
\lim_{n \to \infty} \log \|f\|_{1/n} \le \int_X \log|f| \,d\mu.
$$
Apply the squeeze theorem with lemma 2 to obtain
$$
\lim_{n \to \infty} \log \|f\|_{1/n} = \int_X \log|f| \,d\mu.
$$
Since $\log$ is continuous, we conclude
$$
\lim_{n \to \infty} \|f\|_{1/n} = \exp\left(\int_X \log|f| \,d\mu\right).
$$
The above argument applies to any sequence $s_n$ that converges to $0$, not just $1/n$. The general result stated at the beginning now follows. This is a standard argument in measure theory. It is usually used to apply the dominated convergence theorem to general limits, not just limits of countable sequences.
To answer your other questions, the "scaled norm" follows from the general case as I explained at the beginning of my answer. I've never seen the geometric mean called $L^0$. As for further readings, check out Rudin's Real and Complex Analysis or Folland's Real Analysis. The above is an exercise in one of them (I think the former).
