The answer is $5$. For the example consider the following table (see the cover photo of the booklet).
$$a_{ij} = \begin{cases} 2 & i \equiv 1 \pmod{8} \\ 5 & i \equiv 4, 5 \pmod{8} \\ 4 & i \equiv 0 \pmod{8} \text{ or } i \equiv 7 \pmod{8} \text{ and } j \not\equiv 2, 5 \pmod{6} \\ 1 & i \equiv 3 \pmod{8} \text{ and } j \not\equiv 2, 5 \pmod{6} \\ 3 & \text{otherwise.} \end{cases}$$
Where $a_{ij}$ is the number written in the cell $(i, j)$.

To prove that $5$ is maximum, note that if there is one $8$ in this table all other cells should be $8$. If there is a $7$ in this table then it has a neighbor $a$ which is not $7$, and $a$ should have a neighbour $b$ which is not $7$. Now none of the common neighbors of $a, b$ can be $7$. But it is easy to see that $a, b$ has a common neighbor with number $7$ which is a contradiction. Hence, $7$ can not appear in this table.

Assume that all the numbers $\{1, 2, 3, 4, 5, 6\}$ will appear in this table. Define the taxi-cab distance between $(i, j)$ and $(i', j')$ to be $|i - i'| + |j - j'|$. Let $a_{i',j'} = 5$ and $(i,j)$ be a cell with number $6$ with minimal taxi-cab distance with $(i', j')$. We claim that $|i - i'| + |j - j'| = 1$ which is equivalent to say that $(i, j)$ and $(i', j')$ have a common edge.

To prove the claim, note that if $i \neq i'$ and $j \neq j'$, because of symmetry we may assume $i > i'$, $j > j'$. Note that we have $a_{i+1,j} = 6$ or $a_{i,j+1} = 6$ or $a_{i+1,j+1} = 6$ and all these three cells have less taxi-cab distance with $(i', j')$ than $(i, j)$. So every cell filled with number $6$ having minimal taxi-cab distance to $(i', j')$ should be in the same row or column with this cell. Suppose that $j = j'$ and $i < i'$. Now, we have $a_{i+1,j+1} = 6$ or $a_{i+1,j} = 6$ or $a_{i+1,j-1} = 6$. But all three cases are impossible. Indeed $a_{i+1,j} = 6$ contradicts minimality of taxi-cab distance from $(i,j)$ to $(i',j')$, and $a_{i+1,j \pm 1} = 6$ contradicts the observation above that cells with number $6$ and minimal distance to $(i',j')$ should be in the same row or column. So the claim is proved.

Now consider the two adjacent cells with numbers $6,5$. These two cells have $10$ neighbors in total, $6$ of them should be $6$ and $5$ of them should be $5$ which is impossible. Thus, the maximum possible number is $5$ and the solution is complete.