A typical $L^p$ function does not have a well-defined trace on the boundary

The key is that you can not find a bounded operator. Such an Operator is continuous. So what happen when you plug in a sequence of continuous functions?

Assume $\Omega=(0,1)$ and choose $$f_n (x):=\begin{cases} 1 &\text{ on } [0,1-1/n]\\ n-nx &\text{ on } [1-1/n,1]\end{cases}$$ So $f_n$ falls linearly to zero on $[1-1/n,1]$.

Each $f_n$ is a continuous function and has trace zero in $1$. Now we restrict the trace operator to $x=1$. Now $0=\lim_n T(f_n)\neq T(f)=1$ since $f_n\to 1$ in $L^2$.

For completeness, I'll expand the idea in this comment. The construction does not require $C^1$ boundary, and works in every bounded domain. Let $$ u_n(x) = (1-n\operatorname{dist}(x,\partial U))^+$$ which is a continuous function on $\overline{U}$. Since the sequence $u_n^p$ is decreasing, it is dominated by $u_1^p$, which is integrable. Hence $$ \lim_{n\to\infty}\int_U u_n^p = \int_U \lim_{n\to\infty} u_n^p = 0 $$ On the other hand, $$\int_{\partial U}u_n^p = \int_{\partial U}1\not\to0$$ which yields the claim.