I'll prove the general case in which $\mu$ is a positive measure on a space $X$ and $\mu(X) = 1$. 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 $\log a \le a - 1$ with $a = \int_X |f|^{1/n} \,d\mu$ 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 $\lim_{x \to 0} \dfrac{a^x - 1}{x} = \log a$. Take the limit of (1) as $n \to \infty$ and 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:
$$
\lim_{n \to \infty} \|f\|_{1/n} = \exp\left(\int_X \log|f| \,d\mu\right)
$$
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).
