This response is closely related to my answer Here.
For the case $W^{2,n}$ we automatically get Holder regularity by applying Sobolev and then Morrey. However, we don't get Lipschitz. The following example "integrates" the counterexample to boundedness of $W^{1,n}$ functions.
Take a function $\psi$ supported on $B_2$ with $\psi$ linear and nonconstant on $B_1$ and $|\nabla \psi| < c$, and add dyadic rescalings together as follows. Consider $$u(x) = \sum_{i=1}^{\infty} h_i\psi(2^ix)$$ for some $h_i$ we will choose to get bounded $W^{2,n}$ norm but unbounded derivative. Note that $|D^2(h_i\psi(2^ix))|$ grows like $h_i2^{2i}$ and they are supported on disjoint dyadic rings of volume going like $2^{-in}$. Thus, to get bounded $W^{2,n}$ norm we want $$\sum_i h_i^n2^{in} < C.$$
To give unboundedness of the derivative we want $$\sum_i h_i2^{i} = \infty.$$
The natural choice for $h_i$ is $\frac{2^{-i}}{i}$, which gives the counterexample.
Remark: This function has size $~ 2^{-k}\sum_{i=1}^k \frac{1}{i}$ ~ $2^{-k}|\log\log(2^{-k})|$ at $|x| = 2^{-k},$ so a more explicit example might look something like $|x||\log\log(|x|)|$, which looks almost like a cone away from $0$ but the slope gets unboundedly high near $0$.
