Consider the standard mollifier $$ \phi(x) = C\exp\left(-\frac{1}{1-x^2}\right), \quad -1<x<1. $$ such that $\int\phi(x) = 1$.

Let $f(x) = |x|$ and consider the convolution $f\ast \phi$. I am interested in a bound in the derivatives $(f\ast\phi)^{(k)}$, say, how large is $\max_{x\in [-1,1]} |(f\ast\phi)^{(k)}|$?

Question 1. What is an upper bound on $\|\phi^{(k)}\|_\infty$? I think this is $C_1(C_2k)^{2k} \cdot k^{C_3}$ but I didn't seem to find it in the literature. Can anyone point me to some reference?

Question 2. Is there a mollifier (not necessarily compactly supported) such that $\int \phi(x) = 1$ and $\|\phi^{(k)}\|_\infty \leq C_1(C_2k)^k k^{C_3}$? (It seems that $k^{2k}$ is the common bound for several mollifiers, see this post on stackexchange and the appendix in this note by Ben Green. I'd like to know if it's possible to have just a bound of $k^k$.)

Question 3. Is there a bound for $\|\phi^{(k)}\|_1$? The trivial bound $\|\phi^{(k)}\|_1\leq 2\|\phi^{(k)}\|_\infty$ is absolutely too cruel because the spike for the bad $\ell_\infty$ norm is very narrow. For instance, $\|\phi^{(4)}\|_\infty\approx 8000$ while numerical integration gives that $\|\phi^{(4)}\|_1 \approx 1076.13$.

Question 4. Here $f(x) = |x|$, the cruel bound $|f\ast \phi^{(k)}| \leq \max_{x\in [-1,1]} |f(x)| \cdot \|\phi^{(k)}\|_1$ seems too loose. While $\|\phi^{(4)}\|_1 \approx 1076.13$, the true $\max_{x\in [-1,1]} |(f\ast\phi)^{(k)}|\approx 15$. What is a tighter bound in general?