What you are asking for is not possible, not even if you ask for passive security. Here is a sketch of a proof.
Suppose for sake of contradiction you have an $n$-party MPC protocol $\Pi$ for $f(x_1, \ldots, x_n) = x_1 \land x_n$, where the $x_i$'s are bits. This protocol is secure against a computationally unbounded adversary who can passively corrupt $\ge n/2$ parties.
You can take this $n$-party protocol $\Pi$ and construct a related 2-party protocol $\Pi^*$. Player 1 in $\Pi^*$ plays the role of parties 1 through $n/2$ in $\Pi$, and Player 2 in $\Pi^*$ plays the role of parties $n/2+1$ through $n$ in $\Pi$. So $\Pi^*$ is a 2 party protocol that takes input $x_1$ for Party 1 and $x_n$ for Party 2 and computes $x_1 \land x_n$.
The 2-party protocol $\Pi^*$ is secure against 1 corrupt party. Corrupting 1 party in $\Pi^*$ is like simultaneously corrupting $n/2$ parties in $\Pi$, and by our assumption $\Pi$ is secure in that scenario.
So now we have a 2-party protocol $\Pi^*$ for securely computing the AND of two bits, which is secure against a passive, computationally unbounded adversary (who corrupts 1 party). But this is known to be impossible. If you're asking about perfect security, it was proven in:
The proof was later generalized to statistical security in: