Does $\bigcap_{n=1}^{+\infty}(-\frac{1}{n},\frac{1}{n}) = \varnothing$?
No. It is not empty. Since $0\in (-1/n, 1/n)$ for all $n$, so $0\in\bigcap_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)$.
No, it is not empty, but in fact, if $I_n$ is closed and $I_{n+1}\subseteq I_{n}$, that is not enough to ensure that $$\bigcap_{n=1}^\infty I_n\neq\varnothing.$$ As an example, take the closed intervals $[n,\infty)$. In order for what you have written to hold true, one needs that at least one of the $I_n$ is bounded.
Finally, as others have pointed out, the intersection of open sets you have is nonempty since $0\in\left(-\frac{1}{n},\frac{1}{n}\right)$ for all $n\in\mathbb{N}$. As a counterexample, one can consider $$\bigcap_{n=1}^\infty\left(0,\frac{1}{n}\right).$$
As you already have a bunch of nice answers to your question, I just wanted to comment on the relation with the theorem you have referred to. Note that $J_n\subset I_n$ where $J_n = [-\frac{1}{2}n,\frac{1}{2}n]$ and thus we know that $\bigcap\limits_n J_n$ is not empty. However, $\bigcap\limits_n J_n\subset \bigcap\limits_n I_n$ and the latter is hence non-empty as well.
NO!
$$\bigcap_{n=1}^{+\infty}\,\left(-\frac{1}{n},\frac{1}{n}\right)\,\neq\, \varnothing$$
WHY NOT?:
Note that as $\,n \to \infty,\,$ the endpoints of the intervals get increasingly close to $\,0$, but never reach $0$, hence every interval contains $0\,$:
$$\;0 \in \left(-\dfrac1n, \cfrac1n\right)\;\; \forall n \in \mathbb{N}.$$
Indeed,
$$\bigcap_{n=1}^{+\infty}\,\left(-\frac{1}{n},\frac{1}{n}\right) = \{0\}.$$
You might be interested to know that, e.g., $$\bigcap_{n=1}^{+\infty}\,\left(0, \frac1n\right)\,=\, \varnothing$$ It' simply a matter of choosing the correct endpoints of the open intervals.