Proof that $(L_{all})^C$ is not recursively enumerable
Let HALT be the following version of the halting problem: Given a Turing machine $T$, determine whether it halts on the empty input.

Here is a computable reduction from HALT to $L_{all}$: Given a Turing machine $T$, construct a Turing machine $T'$ which erases its input and then transfers control to $T$. You can check that $T \in \mathrm{HALT}$ iff $T' \in L_{all}$. This shows that $L_{all}$ cannot be co-r.e. Indeed, if $L_{all}$ were co-r.e. then HALT would be co-r.e. Since HALT is also known to be r.e., it would be recursive; but we know that HALT is not recursive.

Here is a computable reduction from HALT to the complement of $L_{all}$: Given a Turing machine $T$, construct a Turing machine $T'$ which on input $n$ runs $T$ on the empty input for $n$ steps; if $T$ halted, $T'$ enters an infinite loop, and otherwise $T'$ halts. You can check that $T \in \mathrm{HALT}$ iff $T' \notin L_{all}$. This shows that $L_{all}$ cannot be r.e. Indeed, if $L_{all}$ were r.e. then HALT would be co-r.e., and we get a contradiction like in the preceding paragraph.

In fact, $L_{all}$ is $\Pi_2$-complete; it is sometimes known as TOT or TOTAL. Thus $L_{all}$ is stronger than HALT, which is only $\Sigma_1$-complete.