$\newcommand{\R}{\mathbb{R}}$ The answer is yes. Indeed, let $e=(1,0,\dots,0)\in\R^n$ and $f_2:=D^{2e}f$. Then $|f|\le M$ and $|f_2|\le M_2$ for some real $M,M_2\ge0$, and \begin{equation} \int f\,D^{2e}\phi=\int f_2\phi \end{equation} for all $\phi\in C^\infty_c(\R^n)$. For all $(x_1,\dots,x_n)\in\R^n$, let \begin{equation} f_1(x_1,\dots,x_n):=\int_0^{x_1}f_1(t,x_2,\dots,x_n)\,dt, \end{equation} with the common convention $\int_0^{x_1}:=-\int_{x_1}^0$ for $x_1<0$. Integrating by parts with respect to the first argument, we see that $f_1$ is the weak derivative $D^e f$ of $f$. Since $f_1(x_1,\dots,x_n)$ is continuous in $x_1$, in fact $f_1(x_1,\dots,x_n)$ is the true partial derivative of $f(x_1,\dots,x_n)$ in $x_1$. It suffices to prove

Claim: $|f_1|\le M+M_2/2$.

Proof. Fixing here arbitrary values of $(x_2,\dots,x_n)$, we reduce the consideration to the case $n=1$. Then, by an appropriate version of Taylor's theorem, for any real $x$ and some $c_\pm=c_\pm(x)\in[-1,1]$ we have \begin{equation} f(x\pm1)=f(x)\pm f_1(x)+c_\pm M_2/2, \end{equation} whence, by subtraction, \begin{equation} |f_1(x)|\le\tfrac12|f(x+1)-f(x-1)|+M_2/2\le M+M_2/2, \end{equation} as desired. $\Box$.

$\newcommand{\R}{\mathbb{R}} \newcommand{\al}{\alpha}$ This is a partial answer: Assuming additionally that $D^\al f$ exists in $C(\R^n)$ for $|\al|=1$, let us show that $D^\al f$ is bounded. Indeed, without loss of generality $\al=e:=(1,0,\dots,0)\in\R^n$. We have $|f|\le M$ and $|D^{2e}f|\le M_2$ for some real $M,M_2\ge0$.

By fixing arbitrary values of the last $n-1$ arguments of the functions, we reduce the consideration to the case $n=1$, so that $D^e f=f'$, the derivative of $f$. Then, by an appropriate version of Taylor's theorem, for any real $x$ and some $c_\pm=c_\pm(x)\in[-1,1]$ we have \begin{equation} f(x\pm1)=f(x)\pm f'(x)+c_\pm M_2/2, \end{equation} whence, by subtraction, \begin{equation} |f'(x)|\le\tfrac12|f(x+1)-f(x-1)|+M_2/2\le M+M_2/2, \end{equation} so that $|f'|\le M+M_2/2$.