http://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm
.$\quad$1. Can the independence assumption be weakened, similar to this?
.$\quad$2. Can the identically distributed assumption be dropped/weakened, in the latter case similar to this?
.$\quad$3. Can the result be finetuned, presumably to something of the form
$\displaystyle\limsup_{n\to \infty} \frac{\frac{S_n}{\sqrt{2\cdot n\cdot \operatorname{log}(\operatorname{log}(n))}}1}{f(n)} \; \; $ some_relation_symbol $\;$ some_constant $\qquad$ almost surely $\qquad \; \; $ ?



For your third question, see a paper of Erdös, where he proves an even more precise result (at least, in the special case $S_n=\sum_{i=1}^nY_i$ where $Y_i$ (independent) are $\pm 1$ with probability $\frac 12$), namely that for $\delta>0$, the following holds with probability one: $$S_n>\left(\frac{2n}{\log\log n}\right)^{1/2}(\log\log n+\frac 34\log\log\log n+\frac 12\log\log\log\log n+\cdots+(\frac 12\delta)\log^{(k)}n)\qquad\text{for infinitely many $n$}$$ and $$S_n>\left(\frac{2n}{\log\log n}\right)^{1/2}(\log\log n+\frac 34\log\log\log n+\frac 12\log\log\log\log n+\cdots+(\frac 12+\delta)\log^{(k)}n)\qquad\text{for only finitely many $n$}$$ In particular, this implies that: $$\limsup_{n\to \infty} \frac{\frac{S_n}{\sqrt{2n\log\log n}}1}{\frac{\log\log\log n}{\log\log n}}=\frac 34$$ 


For your 1st question. Yes. See e.g. "Note on the law of the iterated logarithm for stationary processes satisfying mixing conditions" by H. Oodaira and K. Yoshihara in Kodai Math. Sem. Rep. Volume 23, Number 3 (1971), 335342.. For strong mixing processes it was shown in M. Kh. Reznik, “The law of the iterated logarithm for some classes of stationary processes”, Teor. Veroyatnost. i Primenen., 13:4 (1968), 642–656. 

