I know this is equivalent to showing $\dfrac{f_1 + f_2 + \cdots + f_n} n - f \to 0$.
And in turn $\dfrac{(f_1 - f) + (f_2 - f) + \cdots + (f_n - f)}{n} \to 0$.
I can split the fraction as follows:
$$\frac{(f_1 - f) + (f_2 - f) + \cdots + (f_N - f)}{n} + \frac{(f_{N+1} - f) + (f_{N+2} - f) + \cdots + (f_n - f)} n \to 0.$$
In the first fraction, the numerator differences are bounded, and there are finite terms, so it goes to zero. I am unsure how to prove the remaining $n-N$ terms in the second fraction are bounded, though.
Is it as simple as saying that because we know that $f_n \to f$ uniformly on $S$, we can choose $N$ large enough such that each difference will be less than $\frac{\epsilon}{n-N}$, and the sum of the $n-N$ remaining difference terms will be less than $\epsilon$?
Given $\varepsilon > 0$, there exists an index $N\in\mathbb{N}$ such that $$ |f_j(x) - f(x)| < \varepsilon \qquad \forall j>N,\ \forall x. $$ If $C>0$ is a constant such that $|f_j(x)| \leq C$ for every $j$ and $x$, using your notation you get, for every $n>N$, $$ \frac{1}{n}\sum_{j=1}^n |f_j - f| \leq \frac{C\, N}{n} + \frac{1}{n}\sum_{j=N+1}^n |f_j-f| \leq \frac{C\, N}{n} + \varepsilon \frac{n-N}{n} $$ hence $$ 0\leq \limsup_n \frac{1}{n}\sum_{j=1}^n |f_j - f| \leq \varepsilon. $$ Finally, since this inequality holds for every $\varepsilon> 0$, you conclude that $\frac{1}{n}\sum_{j=1}^n |f_j - f|$ converges uniformly to $0$.