1. **Define the sets $S_j$:**
For each integer $1 \le j \le 2017$, the set $S_j$ consists of integers $0 \le i \le 2^{2017} - 1$ such that $\left\lfloor \frac{i}{2^{j-1}} \right\rfloor$ is an odd integer. This can be written as:
$$S_j = \{ i \mid 0 \le i \le 2^{2017} - 1, \left\lfloor \frac{i}{2^{j-1}} \right\rfloor \text{ is odd} \}$$

2. **Express the polynomial $P$:**
The polynomial $P$ is given by:
$$P(x_0, x_1, \ldots, x_{2^{2017} - 1}) = \prod_{1 \le j \le 2017} \left(1 - \prod_{i \in S_j} x_i \right)$$
Let $T_j = \prod_{i \in S_j} x_i$. Then:
$$P(x_0, x_1, \ldots, x_{2^{2017} - 1}) = \prod_{1 \le j \le 2017} (1 - T_j)$$

3. **Evaluate the polynomial $P$:**
The polynomial $P$ evaluates to 1 if all $T_j$ are 0, and 0 if at least one $T_j$ is 1. We need to count the number of tuples $(x_0, x_1, \ldots, x_{2^{2017} - 1}) \in \{0, 1\}^{2^{2017}}$ such that all $T_j$ are 0.

4. Use complementary counting and the Principle of Inclusion-Exclusion (PIE):
Let $A_j$ denote the set of tuples where $T_j = 1$. We want to compute:
$$X = |P| - \sum |A_j| + \sum |A_i \cap A_j| - \sum |A_i \cap A_j \cap A_k| + \cdots$$
where $|P| = 2^{2^{2017}}$.

5. Calculate the sizes of intersections:
- $|A_j| = 2^{2^{2016}}$ because $|S_j| = 2^{2016}$.
- For $|A_i \cap A_j|$ with $i < j$, $S_j$ fills half of the gaps in $S_i$, so $|A_i \cap A_j| = 2^{2^{2015}}$.
- Generally, $|A_{a_1} \cap A_{a_2} \cap \cdots \cap A_{a_k}| = 2^{2^{2017 - k}}$.

6. Sum using PIE:
$$X = \sum_{k=0}^{2017} (-1)^k \binom{2017}{k} 2^{2^{2017 - k}}$$
Simplifying modulo 2017:
$$X \equiv 2^{2^{2017}} - 2 \pmod{2017}$$

7. **Compute $2^{2^{2017}} \mod 2017$:**
Using Fermat's Little Theorem, $2^{2016} \equiv 1 \pmod{2017}$. We need to find $2^{2017} \mod 2016$:
$$2^{2017} \equiv 2 \pmod{2016}$$
Thus:
$$2^{2^{2017}} \equiv 2^2 = 4 \pmod{2017}$$

8. Final calculation:
$$X \equiv 4 - 2 = 2 \pmod{2017}$$

The final answer is $\boxed{2}$