$\def\low{\mathop{\mathrm{low}}}\def\perm{\mathop{\mathrm{perm}}}$This seems to be a (hopefully correct now) answer to Q1 and Q2. But this looks a bit strange --- perhaps, it is worth it to check some consequences on small lattices.
We implement Birkhoff's theorem in a dual form, identifying $L$ with the lattice of upper sets of some poset $P$ (the structure of which is known); we thus regard each $x\in L$ as an upper set in $P$, so that $x\leq y\iff y\subseteq x$ (!). Next, each element $x\in L$ is determined by the set $\min x$ of its minimal elements, and each independent set $Q\subseteq P$ determines an element $u_Q=\{x\in P\colon \exists q\in Q\quad x\geq q\}$ with $\min u_Q=Q$.
1. First, let us learn the structure of $C^{-1}=(\mu(x,y))$. As has been already mentioned in the comments, $\mu$ is the Möbius function of $L$. Fix an element $x$; for every $J\subseteq \min x$, introduce $x^J=x\setminus J$. Clearly, all the $x^J$ are pairwise distinct. Then one can easily see that $\mu(x,x^J)=(-1)^{|J|}$, and these are the only nonzero values of $\mu(x,\cdot)$ (indeed, the matrix $C'$ determined by these values satisfies $C'C=I$.
(Not needed) Similarly, denoting $\low y=\max(P\setminus y)$, we may define $y_J=y\cup J$ for any $J\subseteq \low y$ and see that $\mu(y_J,y)=(-1)^{|J|}$ are the only nonzero value of $\mu(\cdot,y)$.
2. Let now $-G_L=C^{-1}C^T=(g_{xy})$ (we omit the minus sign for clarity; this changes the sign of the permanent in a clear manner). We have
$$
g_{xy}=\sum_{z\geq x\vee y}\mu(x,z)
=\sum_{z\subseteq x\cap y}\mu(x,z)
=\sum_{\textstyle{J\subseteq \min x\atop
x\subseteq J\cup y}}\mu(x,x^J)
=\begin{cases}
(-1)^{|\min x|}, & y\cap x=x\setminus \min x;\\
0, & \text{otherwise.}
\end{cases}
$$
To state it simple, the element $g_{xy}$ is nonzero if and only if $y\cap x=x\setminus \min x$, and all such elements for a fixed $x$ are the same, namely, they equal $(-1)^{|\min x|}$.
This already yields that all nonzero summands in $\perm G_L$ are equal --- namely, each of them equals $(-1)^{\sum_x|\min x|}$. This shows that the answer to Q1 is affirmative if and only if the permanent has a unique nonzero summand (which is what I called strange), and, given that, answers Q2. We are now to check that strange claim.
3.
So, we need to show that there exists a unique permutation $\sigma\colon L\to L$ satisfying $x\cap \sigma(x)=x\setminus\min x$. We start with constructing such permutation $\tau$, and then show its uniqueness.
For any $x\in L$, let
$$
\tau(x)=\bigcup_{\textstyle{y\in L\atop y\cap x=x\setminus\min x}}y;
$$
that is, $\tau(x)$ is the minimal (in $L$) element satisfying the required property. We show that $\tau$ is a permutation by indicating its right inverse $\tau^{-1}$ as
$$
\tau^{-1}(y)=u_Q,\qquad \text{where } Q=\max(P\setminus y).
$$
Indeed, it is clearly seen that $\tau(\tau^{-1}(y))=y$ for all $y\in L$.
Finally, let $\sigma$ be a permutation satisfying the conditions above; then $\sigma(x)\subseteq \tau(x)$ for all $x\in L$. Then
$$
\sum_{y\in L}|y|=\sum_{x\in L}|\sigma(x)|\leq \sum_{x\in L}|\tau(x)|=\sum_{y\in L}|y|,
$$
so the middle inequality turns into equality. This may happen only if $\sigma=\tau$, which finishes the proof.
