No, this already fails in $\mathbb{R}^1$. Let $K_0$ be the middle thirds Cantor set and define $f_0(t) = {\rm dist}(t,K_0)$. Then $f_0$ clearly has Lipschitz number $1$ but it is not differentiable at any point of $K_0$. (Since it is nonnegative and vanishes on $K_0$, the only possible value of the derivative at any point of $K_0$ is zero, but for any point in $K_0$ and any $n \in \mathbb{N}$ there will be a point at most $\frac{3}{2} 3^{-n}$ units away on which $f_0$ takes the value $\frac{1}{2} 3^{-n}$.)

Now let $K_1$ be $K_0$ with a scaled-down copy of the Cantor set inserted in each gap. Then the function $f_1(t) = {\rm dist}(t,K_1)$ again has Lipschitz number $1$ and is not differentiable anywhere on $K_1$, for the same reason. Well, I suppose the distance function from any closed measure zero set $A$ would not be differentiable anywhere on $A$. Anyway, continue in this manner and define $f = \sum 2^{-n}f_n$.

This function has Lipschitz number at most $2$, and it clearly is not differentiable at any point of $K_0$ because, in the argument which showed this for $f_0$, it now takes an even larger value at the nearby point. For the same reason, for any $n$ the function $\sum_{i=n}^\infty 2^{-i}f_i$ is not differentiable at any point of $K_n\setminus K_{n-1}$, and therefore neither is $f$, because $f$ differs from this sum by the function $\sum_{i=0}^{n-1}2^{-i}f_i$ which is linear in a neighborhood of the point. We conclude that $f$ is Lipschitz but not differentiable at any point of the meager set $\bigcup K_n$.

No, not even in one dimension. Googling "not differentiable on a residual set" yields a citation of the paper F. Mignot, Contrôle dans les inéquations variationelles elliptiques, J. Functional Analysis 22 (1976), 130–185.