Answer: 640

Small Hint:

Each entry feeds upward like Pascal’s triangle: for bottom entries $x_0,$ $\ldots,$ $x_{10},$ the top square is $\sum_i \binom{10}{i} x_i$

Big Hint:

Modulo $3,$ except for $i = 0,$ $1,$ $9,$ $10,$ the coefficient $\binom{10}{i}$ vanishes, so only $x_0 + x_1 + x_9 + x_{10}$ matters

Solution:

Label the bottom-row entries $x_0,$ $x_1,$ $\ldots,$ $x_{10}.$ Since each square is the sum of the two below it, the contributions accumulate with Pascal’s-triangle weights: the top square equals $\sum_{i=0}^{10} \binom{10}{i} x_i.$

Modulo $3,$ direct checking (or Lucas’ theorem with $10 = 101_3$) shows that for $2 \le i \le 8,$ $\binom{10}{i} \equiv 0,$ while $\binom{10}{0} = \binom{10}{10} = 1$ and $\binom{10}{1} = \binom{10}{9} = 10 \equiv 1.$ So the top square is a multiple of $3$ exactly when $x_0 + x_1 + x_9 + x_{10}$ $\equiv 0 \pmod 3.$

Because the four entries are $0$ or $1,$ their sum is divisible by $3$ only when it is $0$ or $3:$ either all four are $0$ (one way) or exactly three are $1$ (four ways), for $5$ choices. The remaining seven entries $x_2,$ $\ldots,$ $x_8$ are free, so the count is $5 \cdot 2^7 = 640.$