When $0$ or $1$ is written on each tile in such a way that the product of the two numbers written on every neighboring pair of tiles is always $0$, we'll call the status a z-pattern. First, we prove a couple of lemmas.

Lemma 1. Let $a_n$ be the number of z-patterns for $n$ tiles laid in a row. Then
$$a_n = F_{n+2}$$
for all $n \ge 1$, where $F_n$ is the $n$-th Fibonacci number defined by $F_1 = 1$, $F_2 = 1$, and $F_{n+2} = F_{n+1} + F_n$ for all $n \ge 1$.

Proof. We prove by induction on $n \ge 1$. It is clear that the lemma holds for $n=1$ and $n=2$. Assume the lemma for all $k$, $1 \le k < n$, where $n \ge 3$. If $0$ is written on the tile at one end, then any number ($0$ or $1$) can be written on the next tile and hence there are $a_{n-1}$ such z-patterns. If $1$ is written there, then $0$ should be written on the next tile and hence there are $a_{n-2}$ such z-patterns. Thus we get
$$a_n = a_{n-1} + a_{n-2} = F_{n+1} + F_n = F_{n+2},$$
which proves the lemma.

Lemma 2. Let $b_n$ be the number of z-patterns for $n$ tiles fixed on a wall in a ring shape. Then
$$b_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1}$$
for all $n \ge 4$.

Proof. Let $n \ge 4$ and choose any tile among the $n$ tiles fixed on a wall in a ring shape. If $0$ is written on the tile, then any number can be written on the neighboring tiles and hence there are $a_{n-1}$ such z-patterns. If $1$ is written on the tile, then $0$ should be written on the neighboring two tiles and hence there are $a_{n-3}$ such z-patterns. Thus we get
$$b_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1},$$
which proves the lemma.

We now return to the problem. Consider the $2 \times 2$ tiles in the center. Observe that the number of $0$'s that can be written on these four tiles equals $4$, $3$ or $2$.

Case 1) Four $0$'s:
$$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$
We may apply Lemma 2 to the $12$ tiles surrounding the four tiles in the center because any number can be written on the $12$ tiles. Therefore, the number of z-patterns in this case equals
$$b_{12} = F_{13} + F_{11} = 233 + 89 = 322.$$

Case 2) Three $0$'s:
$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$
$$\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$$
$$\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$
In each of the four subcases, only $0$ can be written on two neighboring tiles of the tile marked by $1$. For the remaining $10$ tiles, it is clear that the number of z-patterns equals to $a_9 \times 2 = F_{11} \times 2 = 178$. Therefore, the total number of z-patterns in this case equals
$$178 \times 4 = 712.$$

Case 3) Two $0$'s:
$$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
In each of the two subcases, only $0$ can be written on four neighboring tiles of the two tiles marked by $1$. For the remaining $6$ tiles, it is clear that the number of z-patterns equals to $a_3^2 \times 2^2 = F_5^2 \times 4 = 100$. Therefore, the total number of z-patterns in this case equals
$$100 \times 2 = 200.$$

Combining the three cases, the answer is: $322 + 712 + 200 = 1234$. $\square$