It would be nice if the rule for determinants for $2\times2$ matrices generalized to the case of $2n\times 2n$ matrices:
$\det \begin{pmatrix}
A & B \cr
C & D
\end{pmatrix}
=\det A \det D - \det B\det C$,
but this is sadly not true.
Nonetheless, the familiar Laplace expansion theorem for minors of order $n-1$ does have a generalization to minors of any order, including, in this case, minors of order $2n$ of a $4n \times 4n$ matrix, see
http://www.proofwiki.org/wiki/Laplace's_Expansion_Theorem
This might help.
Let's start by constructing an eigendecomposition of $L_n$.
Let $\mathbf{v}_j \in \mathbb{R}^n$, for $1 \leq j \leq \lfloor (n-1)/2 \rfloor$, be the vector defined as follows:
- its $j$-th entry is $\sqrt{\frac{2j - n}{2j - n - 1}}$;
- its $k$-th entry (for $j+1 \leq k \leq n-j$) is $\frac{-1}{\sqrt{(2j - n)(2j - n - 1)}}$;
- all other entries are $0$.
Similarly, let $\mathbf{v}_j \in \mathbb{R}^n$, for $n + 1 - \lfloor n/2 \rfloor \leq j \leq n$ be defined by:
- its $j$-th entry is $\sqrt{\frac{2j - n - 1}{2j - n}}$;
- its $k$-th entry (for $n - j + 1 \leq k \leq j-1$) is $\frac{-1}{\sqrt{(2j - n)(2j - n - 1)}}$;
- all other entries are $0$.
Finally, define $\mathbf{v}_j \in \mathbb{R}^n$ for $j = n - \lfloor n/2 \rfloor$ to be the zero vector. We have now defined $\mathbf{v}_j$ for all $1 \leq j \leq n$.
These vectors are mutually orthonormal eigenvectors for $L_n$, and the eigenvalue corresponding to eigenvector $\mathbf{v}_j$ is exactly $j$ itself:
$$
L_n = \sum_{j=1}^n j\mathbf{v}_j\mathbf{v}^*_j
$$
From this it immediately follows that the pseudo-inverse $L_n^+$ is
$$
L_n^+ = \sum_{j=1}^n \frac{1}{j}\mathbf{v}_j\mathbf{v}^*_j.
$$
Note that I did some trickery here, since one eigenvalue of $L_n$ is always zero -- instead of setting the eigenvalue to zero, I set the corresponding eigenvector $\mathbf{v}_j$ for $j = n - \lfloor n/2 \rfloor$ to be zero, since this made the final answer cleaner in my opinion.
Also note that I've stated a bunch of things without proof here. They can be proved easily enough in a brute-force sort of way; it's just messy and not terribly enlightening.
Best Answer
Not an answer, but an amusing observation: The determinant of the matrix usually (but not always) has denominator a perfect square (empirically), while the numerator always seems to have a huge prime factor.
EDIT Actually, it seems that the expectation of $\log(P(n))/\log(n),$ where $P(n)$ is the largest prime factor of $n$ approaches the Golomb-Dickman constant, which is about 0.62. In view of this, the largest prime factor of the numerator is pretty much par for the course, or at least, not obviously NOT par for the course.
n=1: det = 1/1
n=2: det=7/(12^2)
n=3 det = 647/(2160^2)
n=4 det = (19 * 571)/(672000^2)
n=5 det = (179 * 179357)/(7*4233600000^2)
n=6 det = (97 * 157 * 384191938531)/(186313420339200000^2)
n=7 det = (23 * 1280587616051046200369)/(2067909047925770649600000^2)
n=8 det = (317 * 6337 * 25997 * 87403 * 511645991608091)/(365356847125734485878112256000000^2)
(and some more. Note that mysterious 7 which keeps popping up in the denominator, for n=5,9,11 )
n=9 det = (55091 * 7731550926975871647518813143593349)/(7 * 146968826339795671126721851844198400000000^2)
n=10 det = (257 * 47360083 * 530916328215423816923887043836865928533)/(15402297982638230438765209613012092908994560000000000^2)
n=11 det = (31 * 1193 * 2647 * 538580971 * 957346850101 * 71222443011485886519799225151)/(7 * 175251348661711183890804992735665222783492739007774720000000000^2))