I offer a proof that gives tighter sup-norm bounds when you have control of the $L^2$ norm. The following proof should also be generalizable to general Holder classes (e.g. Lipschitz derivatives). I give the rigorous proof for $d=1$ as the generalization to dimension $d$ follows easily.


Suppose that $f_n$ and $f_0$ are L-Lipschitz on the interval I:=[A,B]. Let $\varepsilon_n := \|f_n-f_0\|_{\infty, I}$. 

The key idea of the proof is the following proposition.

Proposition (1). Fix $\delta_n := \frac{\varepsilon_n}{4L}$. I claim that there exists some $c \in [A, B - \delta_n]$ such that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \geq \varepsilon_n/2$. We will prove this by proving the contra-positive. To this end, suppose that for all $c \in [A, B - \delta_n]$ that $\text{inf}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \varepsilon_n/2$. Take some $c \in [A, B - \delta_n]$ and let $t_n$ be such that $ |f_n(t_n) - f_0(t_n)| \leq \varepsilon_n/2$. Then,
$$\text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_0(x)| \leq \text{sup}_{x \in [c,c+\delta_n]} |f_n(x) - f_n(t_n)| + |f_n(t_n) - f_0(t_n)| + \text{sup}_{x \in [c,c+\delta_n]} |f_0(t_n) - f_0(x)| \leq 2L\delta_n + \varepsilon_n/2,$$
where the RHS bound holds independent of $c$ and $t_n$. Thus, we actually have
$$\text{sup}_{x \in I} |f_n(x) - f_0(x)| \leq 2L\delta_n + \varepsilon_n /2\leq \varepsilon_n$$
by our choice of $\delta_n$. This proves the claim. (One can prove similar propositions for Holder classes more generally (e.g. Lipschitz derivative.) 

Let $B_n = [c,c+\delta_n]$ be the interval guaranteed by proposition 1 where $\inf_{x \in B_n}|f_n(x) - f_0(x)| \geq  \|f_n-f_0\|_{\infty} =  \varepsilon_n$. It follows that
$$ \varepsilon_n^{2} \lessapprox {\delta_n} \varepsilon_n \lessapprox  {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$ which gives
$$\|f_n - f_0\|_{\infty} \lessapprox  \|f_n - f_0\|_{L^1}^{1/2},$$
and
$$ \varepsilon_n^{3/2} \lessapprox \sqrt{\delta_n} \varepsilon_n \lessapprox  \sqrt{{\int_{B_n} |f_n-f_0|^2(x) dx}} \leq \|f_n - f_0\|_{L^2}$$ which gives
$$\|f_n - f_0\|_{\infty} \lessapprox  \|f_n - f_0\|_{L^2}^{2/3}.$$
In particular, the above shows $L^2$ control gives a better sup-norm bound.



Proposition (1) can be generalized to dimension $d$ with virtually no changes but an addition of a constant $O(\sqrt{d})$. In particular, one can show that $\|f_n-f_0\|_{\infty} = C_1\varepsilon_n$ implies there is a rectangle $B_n$ with sides at most $C_2\varepsilon_n$ such that $\inf_{x\in B_n}|f_n(x)-f_0(x)| \geq C_3 \varepsilon_n$ for constants $C_i$>0. Thus,
$$  \varepsilon_n \varepsilon_n^{ d} \lessapprox {\int_{B_n} |f_n-f_0|(x) dx} \leq \|f_n - f_0\|_{L^1}$$
and
$$  \varepsilon_n \varepsilon_n^{d/2 } \lessapprox \sqrt{\int_{B_n} |f_n-f_0|(x)^2 dx} \leq \|f_n - f_0\|_{L^2}$$
which implies
$$ \|f_n-f_0\|_{\infty}\lessapprox \|f_n - f_0\|_{L^1}^{\frac{1}{1+d}}$$
and
$$  \|f_n-f_0\|_{\infty}  \lessapprox \|f_n - f_0\|_{L^2}^{\frac{2}{d+2}}.$$

If one has that $f_n$ and $f_0$ are univariate Lipschitz with $L$-Lipschitz derivatives, I believe one can show by a similar argument (proving a tighter version of proposition (1) with first order taylor expansions) that
$$\|f_n - f_0\|_{\infty} \lessapprox  \|f_n - f_0\|_{L^1}^{\frac{2}{3}}.$$
$$\|f_n - f_0\|_{\infty} \lessapprox  \|f_n - f_0\|_{L^2}^{\frac{4}{5}}.$$