The terms $x, x + 2, x + 4, \dots, x + 2n$ form an arithmetic sequence, with $x$ an integer. If each term of the sequence is cubed, the sum of the cubes is $-1197$. What is the value of $n$ if $n > 3$?
There are $n+1$ terms in the sequence $x, x+2, x+4, \ldots, x+2n$, and either all of them are even or all of them are odd.  If they were all even, then their cubes would be even and the sum of their cubes would be even.  Therefore, all the terms are odd.

If the sequence contains both positive and negative terms, then it contains more negative terms than positive terms, since the sum of the cubes of the terms is $-1197$.  Also, all of the positive terms will be additive opposites of the first several negative terms, so we may first look for consecutive negative odd numbers whose cubes sum to $-1197$.  If we add cubes until we pass $-1197$, we find that  \[
(-1)^3+(-3)^3+(-5)^3+(-7)^3+(-9)^3=-1225.
\] Since 1197 is 28 less than 1225, we would like to drop two terms than sum to $-28$.  We find that first two terms sum to $-28$, which gives \[
(-9)^3+(-7)^3+(-5)^3=-1197.
\] Filling in negative and positive terms that sum to 0, we find that the possibilities for the original arithmetic sequence are  \begin{align*}
-9, &-7, -5, \text{ and} \\
-9, &-7, -5, -3, -1, 1, 3.
\end{align*} The number of terms is $n + 1$, and $n > 3$, so $n + 1 = 7$, or $n = \boxed{6}$.