It is apparent that $n!\pm k$ are composite and not coprime to $n!$ for $2\le k\le n$. Of course, $n!$ is itself composite and not coprime to itself. Longer runs of composite numbers (up to lengths roughly $2n$) can occur when one or especially both of $n! \pm 1$ happen to be composite.
$n! \pm 1$ are both composite for $n=5,8,9,10,13\dots$; however, $n! \pm 1$ whether or not composite will be relatively prime to $n!$ and such longer runs do not satisfy the requirement in the original question of containing only "not coprime" numbers. For example, the run $114,\dots,126$ identified in a previous answer surrounds $5!=120$ but includes $5!-1=7\cdot 17$ and $5!+1=11^2$. Each of $7,11,17$ is relatively prime to $5!$
By careful choice, examples of $n$ can be manufactured where $n!\pm k$ continues to afford composite, not coprime, values for consecutive integers even when $k>n$. For example, if $n$ is odd, then $n!\pm (n+1)$ will be divisible by $2$ (as will every value of $k=n+2i+1$). We can further identify values of $n$ such that $n$ is odd and $3\mid (n+2)$. The Chinese Remainder Theorem says this process can be repeated to find values of $n$ that meet those requirements and further meet $5\mid (n+4)$, $7\mid (n+6)$, etc. to extend the runs of consecutive numbers not coprime to $n!$ by a few with each step. An example of this kind of extension was given in a previous answer.
What is apparent, however, is that every time we look to make the run of "not coprime" numbers a few longer, the values of $n$ and hence $n!$ grow very rapidly, quickly outstripping our ability to calculate by hand or even with simple calculators.