Recall that the Chinese Remainder Theorem (CRT) states that, for any given integers $r$, $s$ and any two positive co-prime integers $a$, $b$ there exists exactly one integer $x$ that satisfies $0 \le x < ab$, $x \equiv r \pmod a$ and $x \equiv s \pmod b$. We can apply CRT here with $a = 125$ and $b = 8$, because $1000 = 125 \cdot 8$ and $\text{gcd}(125, 8) = 1$.
Because $125$ is odd and $125 < 2019$ it is clear that $n \equiv 0 \pmod{125}$. To calculate $n \pmod 8$, note that $1 \cdot 3 \cdot 5 \cdot 7 \equiv 1 \pmod 8$. Because $n$ is the product of $(2019+1)/2 = 1010$ odd integers and $1010 = 4 \cdot 252 + 2$, we can split the given product as follows
$$n = \prod_{k=0}^{251} (8k + 1)(8k + 3)(8k + 5)(8k + 7) \cdot 2017 \cdot 2019$$
Therefore, $n \equiv (1 \cdot 3 \cdot 5 \cdot 7)^{252} \cdot 1 \cdot 3 \equiv 3 \pmod 8$. Alternatively, when we use the usual notation for odd factorials, $(2k-1)!! = \prod_{j=1}^k (2j-1)$, we can prove by induction on $k \ge 1$ the following table:

When $k=1$ this is obvious. For the inductive step we use that $(2k+1)!! = (2k+1)(2k-1)!!$. Then putting $k=1010$ we have $k \equiv 2 \pmod 4$ and so $n = (2k-1)!! \equiv 3 \pmod 8$.
Having established that $n \equiv 0 \pmod{125}$ and $n \equiv 3 \pmod 8$, to find $n \pmod{1000}$, we note that $n$ is an odd multiple of $125$. We can either check the five odd multiples of $125$ below $1000$ to see which one is congruent to $3$ (mod $8$), or we solve the congruence $125k \equiv 3 \pmod 8$. This is done by first observing that $125 \equiv 5 \pmod 8$ which gives $5k \equiv 3 \pmod 8$, and then multiplying both sides by $5$ to obtain $k \equiv 25k \equiv 5 \cdot 3 \equiv 7 \pmod 8$. Hence, $n \equiv 7 \cdot 125 \equiv 875 \pmod{1000}$, i.e. the last three digits of $n$ are $875$.

Alternatively, we may use the uniqueness in the CRT. It would then be sufficient to check that $875 \equiv 0 \pmod{125}$ and $875 \equiv 3 \pmod 8$.