$\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$ 

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 $f_n\not\to0$. 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 $M$. 
So, for the $f_n$'s, it is impossible get a picture like the one below, where the intervals of significantly non-zero values of $f_n$ become (relatively) negligibly short in time. Therefore, I don't think one can reasonably say that such a sequence has a negligible amount of $1$'s.  
 
[![enter image description here][1]][1]
  


  [1]: https://i.sstatic.net/iS0Ns.png