As the other answer has pointed out, $-1/12$ is not the only value that can obtained with analytic continuation. However, it *is* the unique constant term of the asymptotic expansion of the [smoothed partial sums](https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/), which perhaps explains why it is the most "natural" value.

Let $\eta$ be [any Schwartz function](https://math.stackexchange.com/a/1331028/76284) such that $\eta(0) = 1$. Then

\begin{align}
\sum_{n=0}^\infty n^s \eta(n \varepsilon)
&= \zeta(-s) + O(\varepsilon) + \frac{1}{\varepsilon^{s+1}} \int_0^\infty x^s \eta(x) dx
\end{align}

Therefore, by choosing for any given $s$ an $\eta$ that makes the last integral zero, we get

\begin{align}
\sum_{n=0}^\infty n^s
&= \sum_{n=0}^\infty n^s \lim_{\varepsilon \rightarrow 0^+} \eta(n \varepsilon) \\
&\overset{!}{=} \lim_{\varepsilon \rightarrow 0^+} \sum_{n=0}^\infty n^s \eta(n \varepsilon) \\
&= \lim_{\varepsilon \rightarrow 0^+} \left( \zeta(-s) + O(\varepsilon) \right) \\
&= \zeta(-s)
\end{align}