Bounding supremum norm of Lipschitz function by L1 norm
It can be shown that for $d=1$ the best upper bound on $\|f\|_\infty$ is given by $$\|f\|_\infty\le\sqrt{2L\|f\|_1}\,1(\|f\|_1\le L/2)+(L/2+\|f\|_1)\,1(\|f\|_1>L/2).$$ If there is a sufficient interest, I can later provide details on this.
One can see that, if $L=O(\|f\|_1)$, then the above bound is of the same order of magnitude as the bound $\|f\|_1+L$ given in a comment by Nate Eldredge. On the other hand, the above bound goes to $0$ (as it should) when e.g. $\|f\|_1\to0$ while $L\asymp1$.
We see that the optimal bound is rather complicated already for $d=1$.
The case of $d>1$ is much more complicated -- in particular, because it is hard to evaluate or even estimate the volume of the intersection of the hypercube and an arbitrary Euclidean ball -- cf. e.g. https://math.stackexchange.com/a/2008339/96609.
Anyhow, as Nate Eldredge pointed out, your conjectured bound cannot hold because it should increase with $L$. Also, the bound should of course depend on $\|f\|_1$. So, I think further help depends on whether you can tell us what kind of bound on $\|f\|_\infty$ will suffice for the purposes of your research.
$\newcommand\Om\Omega$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 $L$.