Computational Complexity of Not All Equal SAT variant
Your problem is NP-hard, as can be seen by a (known, I believe) reduction from CNF-SAT: Reduce from (for example) $3$-SAT as follows. Given a formula $F$, create a formula for your special NAE-SAT by introducing exactly one additional variable, say $y$. Add $y$ to every clause in $F$ to obtain $F'$. 

Example: $F = (x_1 \vee x_2 \vee \neg x_3) \wedge (\neg x_2 \vee x_3 \vee \neg x_4)$ becomes  $F' := (x_1 \vee x_2 \vee \neg x_3 \vee y) \wedge (\neg x_2 \vee x_3 \vee \neg x_4 \vee y)$ 

After doing this, every clause contains $y$ and thus shares (at least) one literal.

Proof sketch:
If $F$ is satisfiable, use the same satisfying assignment to satisfy $F'$, together with $y = false$.

If $F'$ is satisfiable, there are two options


$y = false $, in this case the same assignment satisfies $F$ ($F'$ has at least one true literal per clause, and it is never $y$, thus $F$ has at least one true literal per clause as well)
$y = true$, negate the satifying assignment for $F'$ to satisfy $F$ (at least one false literal per clause in $F'$, after negating the assignment we have at least one true literal per clause)