As stated, the answer to your question is no. The Cantor function is a common counterexample when the derivative is required to exist only almost everywhere. It is continuous, has zero derivative a.e., in particular in an open set of full measure, but it is not Lipschitz continuous, nor absolutely continuous. You can make an analogous example in the cube $\mathbb{R}^n$ considering the Cantor function of the first variable.
If instead you take a function $f\in L^1_{loc}(\mathbb{R}^n)$ and require its gradient to exist in the distributional sense and to be represented by a function in $L^\infty$, then you are by definition stating that $f$ belongs to the Sobolev space $W^{1,\infty}(\mathbb{R}^n)$, which can be proven to coincide with the space of Lipschitz functions, see for example this question. This still holds if you replace $\mathbb{R}^n$ with a convex set $\Omega$ (there are some weaker conditions on $\Omega$ under which it works, but it doesn't work with arbitrary domains). Note that you don't need to assume a priori that $f$ is continuous, nor that the gradient is continuous almost everywhere.
Prove that $f_n (x) \to f'(x) $ as $n\to \infty$
For $(1)$ and $(2)$ take $f(x)=x^3$ and show the non-uniform convergence of $f_n$ on $\mathbb{R}$ .
For $(3)$
Claim: $f_n$ converges uniformly to $f'$ on $[0,1]$
Proof : Note that $f$ is continously differentiable, i.e $f'$ is continous .
Since continous function on a compact set is uniformly continous , so $f'$ is uniformly continous on $[0,1]$.
So given $\epsilon \gt 0, \exists \delta \gt 0$ such that $\forall x,y\in [0,1]$ with $|x-y| \lt \delta$ , we have $|f'(x)-f'(y)| \lt \epsilon $
By Archimedan Property of Real Numbers , there exist $N \in \mathbb{N} $ such that $\forall n\gt N$ , have $\frac 1n \lt \delta$
Now ,let $x\in [0,1]$ be arbitary and $n\gt N$
Then , $f(x+\frac 1n)-f(x)=\frac 1n f'(a)$ for some $x \lt a \lt x+\frac 1n$, by Mean Value Theorem
Thus $| f_n(x)-f'(x)|=|f'(a)-f'(x)|\lt \epsilon $ since $|x-a| \lt \frac 1n \lt \delta$
Thus the claim is proved.
$(4)$ is obviously false since the sequence always converges pointwise to the derivative, whatever be the domain.
Best Answer
Every continuously differentiable function is locally lipschitz. However, the function $f(x)=e^x$ is continuously differentiable, but not uniformly lipschitz.
So we are essentially assuming that the derivative exists and is globally bounded.