You did the hard part already.
Let's simplify notation a little. Notice
$$\hat\beta - \beta = (X^\prime X)^{-1}X^\prime (X\beta y + \epsilon) - \beta = (X^\prime X)^{-1}X^\prime \epsilon.$$
Therefore we may write $L_1^2$ as
$$(\hat\beta-\beta)^\prime(\hat\beta-\beta)= \epsilon^\prime X^\prime (X^\prime X)^{-2} X \epsilon = \epsilon^\prime A \epsilon = \sum_{i,j} \epsilon_i\, a_{ij}\, \epsilon_j.$$
Note that $A$ is symmetric: $a_{ij} = a_{ji}$ for all indexes $i$ and $j.$ Moreover,
$$\operatorname{Tr}(A) = \operatorname{Tr}\left(X^\prime (X^\prime X)^{-2} X\right)=\operatorname{Tr}\left(X^\prime X(X^\prime X)^{-2} \right)=\operatorname{Tr}\left((X^\prime X)^{-1}\right)$$
and similarly
$$\operatorname{Tr}(A^2) = \operatorname{Tr}\left((X^\prime X)^{-2}\right).$$
Choose units of measurement for the $y_i$ that make $\sigma^2=1$ so we don't have to track it: we know this will introduce a factor of $\sigma^4$ at the end.
The only fact about Normal variates we will need is that when the $\epsilon_i$ are independent standard Normal variables,
$$E[\epsilon_i\epsilon_j\epsilon_k\epsilon_l] = \delta_{ij}\delta_{kl} + \delta_{ik}\delta_{jl} + \delta_{il}\delta_{kj}$$
where $\delta_{ij} = 1$ when $i=j$ and $0$ otherwise is the Kronecker delta. This scarcely needs proof, because a little reflection on its structure shows it merely states the following:
The expectation is zero unless the $\epsilon$'s can be paired up, because otherwise the symmetry of the standard Normal distribution shows the expectation equals its negative.
When two of the $\epsilon$'s are equal, they introduce a factor of $1$ in the expectation (because they have unit variance).
In the special case where all four of the $\epsilon$'s are equal, we obtain the Normal kurtosis, which is $3.$
To compute the variance, we need to find the expected square, which is accomplished by invoking the foregoing result and the linearity of expectation:
$$\eqalign{
E[((\hat\beta-\beta)^\prime(\hat\beta-\beta))^2] &= E\left[\sum_{i,j}\epsilon_i\, a_{ij}\, \epsilon_j\ \sum_{k,l}\epsilon_k\, a_{kl}\, \epsilon_l\right] \\
&= \sum_{i,j,k,l} a_{ij} a_{kl} \left(\delta_{ij}\delta_{kl} + \delta_{ik}\delta_{jl} + \delta_{il}\delta_{kj}\right) \\
&= \sum_{i,k} a_{ii}a_{kk} + \sum_{i,j}a_{ij}a_{ij} + \sum_{i,k}a_{ik} a_{ki} \\
&=\operatorname{Tr}(A)^2 + 2\operatorname{Tr}(A^2).
}$$
Subtracting off $(E[L_1^2])^2 = \operatorname{Tr}(A)^2$ yields the variance which--in terms of the original unit of measure $\sigma$--is
$$\operatorname{Var}(L_1^2) = 2\sigma^4\operatorname{Tr}(A^2) = 2\sigma^4\operatorname{Tr}\left((X^\prime X)^{-2}\right).$$