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.
The map $f : [0,1] \to \mathbb{R}$, $f(0) = 0$ and $f(x) = x^{3/2} \sin(1/x)$ is differentiable on $[0,1]$ (in particular $f'(0) = \lim_{x \to 0^+} f(x)/x = 0$), but it is not Lipschitz (the derivative $f'(x)$ is unbounded).
Best Answer
Continuous and nowhere Lipschitz
An example is given by the Weierstrass function, which is continuous and nowhere differentiable. This can be justified in two ways:
A Lipschitz function is differentiable almost everywhere, by Rademacher's theorem.
Direct inspection of the proof that the function is nowhere differentiable; the estimates used in the proof also imply it's nowhere Lipschitz ($|f(x+h)-f(x)|$ is estimated from below by $|h|^\alpha$ with $\alpha<1$, for certain $x,h$).
Differentiable and nowhere Lipschitz
There are no such examples: a differentiable function on an interval must be Lipschitz on some subinterval. The following is an adaptation of a part of PhoemueX's answer.
The function $f'$ is a pointwise limit of continuous functions: namely, $$f'(x) = \lim_{n\to\infty} n(f(x+1/n)-f(x))$$ where for each $n$, the expression under the limit is continuous in $x$.
Item 1 implies that $f'$ is continuous at some point $x_0$. This is a consequence of a theorem about functions of Baire class 1: see this answer for references.
Continuity at $x_0$ implies $f'$ is bounded on some interval $(x_0-\delta,x_0+\delta)$. The Mean Value Theorem then implies that $f$ is Lipschitz on this interval.