$\newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$
I don't think such a "reasonable" much broader definition is possible. The reason is that the relative frequency $f_n:=s_n/n$ (where $s_n:=\sum_1^n x_i$) varies very slowly in $n$ when $n$ is large. So, it will take $f_n$ a very long time to significantly depart from whatever value it has attained.
To express this thought rigorously, let us now indeed "focus only on defining $f=0$", as you suggested; for any other value of $f$, we can reason quite similarly. Take some $\ep\in(0,1/2)$ and suppose that $f_M\le\ep$ and $f_N\ge2\ep$ for some natural $M$ and $N>M$; this situation would "reasonably" occur if a "generalized limit frequency" is $0$, but $\lim_n f_n\ne0$. Then $s_M\le M\ep$, $s_N\ge2N\ep$, and hence \begin{equation} N-M\ge s_N-s_M\ge(2N-M)\ep, \end{equation} which implies $N\ge\frac{1-\ep}{1-2\ep}\,M$, and so, \begin{equation} s_N-s_M\ge\frac\ep{1-2\ep}\,M. \end{equation} Note that $s_N-s_M$ is the number of $1$'s in the sequence $(x_n)$ for natural $n\in(M,N]$. Thus, this number of $1$'s -- after the time $M$ (at which $f_M$ was $\ep$-close to $0$) -- needed to get a relative frequency at least $2\ep$ is at least proportional to the time $M$. Therefore, I don't think one can reasonably say that such a sequence has a negligible amount of $1$'s.