$\newcommand\Om\Omega$Now consider the general case of any natural $d$. Here we will give an upper bound on $\|f\|_\infty$ in terms of $\|f\|_1$, $L$, and $d$. This bound will be optimal up to a factor depending only on $d$; as follows from a comment of yours, such factors do not matter to you. The mentioned bound will be exact in the case $d=1$.

Indeed, let $I:=[0,1]$, $\Om:=I^d$,
$$M:=\|f\|_\infty=\max_{x\in\Om}|f(x)|=|f(a)|$$
for some $a=(a_1,\dots,a_d)\in\Om$. Then
\begin{equation}
|f(x)|\ge h_a(x):=(M-L|x-a|)_+=L(r-|x-a|)_+ \tag{1}
\end{equation}
for all $x\in\Om$, where $|x-a|$ is the Euclidean norm of $x-a$, $u_+:=\max(0,u)$, and
$$r:=M/L$$
(assuming $L>0$). So,
$$\frac{\|f\|_1}L=\frac1L\int_\Om|f|\ge\int_\Om dx\,(r-|x-a|)_+
=E(r-R)_+,$$
where
$$R:=\sqrt{\sum_1^d(U_i-a_i)^2}$$
and $U_1,\dots,U_d$ are independent random variables each uniformly distributed on $[0,1]$.

Next,
$$E(r-R)_+=E\int_0^r dv\,1(R<v)=\int_0^r dv\,P(R<v),$$
\begin{align*}
P(R<v)&=P\Big(\sum_1^d(U_i-a_i)^2<v^2\Big) \\
&\ge\prod_1^d P(|U_i-a_i|<v/\sqrt d) \tag{2} \\
&\ge\prod_1^d P(|U_i|<v/\sqrt d) \tag{3} \\
&=\min\Big(1,\frac{v}{\sqrt d}\Big)^d=:Q(v).
\end{align*}
So,
\begin{align*}
\frac{\|f\|_1}L&\ge\int_0^r dv\,P(R<v) \\
&\ge\int_0^r dv\,Q(v) \\
&=g(r):=\left\{\begin{aligned}
\frac{r^{d+1}}{c_d^{d+1}}&\text{ if }r\le\sqrt d, \\
r-\frac{d\sqrt d}{d+1}&\text{ if }r\ge\sqrt d,
\end{aligned}
\right.
\end{align*}
where
\begin{equation*}
c_d:=((d+1)d^{d/2})^{1/(d+1)}.
\end{equation*}
Solving now the inequality $\frac{\|f\|_1}L\ge g(r)$ for $r$ and
recalling that $r=M/L=\|f\|_\infty/L$, we get
\begin{align*}
\|f\|_\infty&\le B_0(\|f\|_1,L) \\
&:=Lg^{-1}\Big(\frac{\|f\|_1}L\Big) \\
&=\left\{\begin{aligned}
c_d L^{d/(d+1)}\|f\|_1^{1/(d+1)}&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\
%\|f\|_1&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\
\|f\|_1+\frac{d\sqrt d}{d+1}\,L&\text{ if }\|f\|_1\ge\frac{\sqrt d}{d+1}\,L.
\end{aligned}
\right.
\end{align*}

**Remark 1:** Obviously, the inequality in (1) will turn into the equality if we choose $f=h_a$ with $a=0$, and then the inequality in (3) will turn into the equality as well. Moreover, the inequality in (2) will change the direction if we replace $v/\sqrt d$ in (2) by $v$.
Therefore, the bound $B_0(\|f\|_1,L)$ is optimal up to a factor depending only on $d$.
It also follows that the bound $B_0(\|f\|_1,L)$ is exact when $d=1$, in which case $B_0(\|f\|_1,L)$ is

exactly the same as the exact upper bound on $\|f\|_\infty$ presented in the other answer of mine on this web page (previously obtained somewhat differently).

**Remark 2:** We have $\|f\|_1\le\|f\|_2$, since the Lebesgue measure on $\Om$ is a probability measure. Also, $B_0(\cdot,L)$ is nondecreasing. So,
$$\|f\|_\infty\le B_0(\|f\|_1,L)\le B_0(\|f\|_2,L).$$

**Remark 3:** One can show that $c_d\le\sqrt{2d}$ for all natural $d$.

**Remark 4:**
It follows from Remark 3 that
\begin{align*}
\|f\|_\infty&\le B_1(\|f\|_1,L) \\
&:=\left\{\begin{aligned}
\sqrt{2d}\, L^{d/(d+1)}\|f\|_1^{1/(d+1)}&\text{ if }\|f\|_1<\frac{\sqrt d}{d+1}\,L, \\
%\|f\|_1&\text{ if }\|f\|_1\le\frac{\sqrt d}{d+1}\,L, \\
(d+1)\|f\|_1&\text{ if }\|f\|_1\ge\frac{\sqrt d}{d+1}\,L.
\end{aligned}
\right.
\end{align*}
Note that $B_1(\|f\|_1,L)$ differs from $B_0(\|f\|_1,L)$ by, at most, a factor depending only on $d$. So, in view of Remark 1, the bound $B_1(\|f\|_1,L)$ is optimal as well up to a factor depending only on $d$.
Note also that the exponent of $\|f\|_1$ in the bound $B_1(\|f\|_1,L)$ is $1/(d+1)$ if $\|f\|_1$ is not too large as compared with $L$, and this exponent is $1$ otherwise.

In view of Remark 2, we also have
$$\|f\|_\infty\le B_1(\|f\|_2,L).$$

**Remark 5:** As shown in Willie Wong's comment, if we had a bound on $\|f\|_\infty$ of the form $C(d)L^a\|f\|_1^b$, then the only possible values for $a$ and $b$ would be $d/(d+1)$ and $1/(d+1)$, respectively. However, in view of Remark 4, it is clear that such a bound on $\|f\|_\infty$ is impossible: the exponent of $\|f\|_1$ cannot be greater than $1/(d+1)$ for values of $\|f\|_1$ not too large as compared with $L$, and this exponent cannot be less than $1$ for values of $\|f\|_1$ large enough as compared with $L$.