In general there's no easy way. But there are a lot of tricks you can use, and sometimes they work.
For this problem, the rational root theorem tells us that if there is a rational root, it must be $\pm 1$. Evidently it isn't either of these, so there are no rational roots, and therefore no factors of the form $n-q$.
So we are looking for two second-degree factors: $$n^4+6n^3+11n^2+6n+1 = (an^2+bn+c)(pn^2+qn+r)$$
But we can see from the coefficient of 1 on the $n^4$ term that $a=p=1$. From the constant term of 1 we have either $c=r=1$ or $c=r=-1$, so it's really
$$n^4+6n^3+11n^2+6n+1 = (n^2+bn\pm1)(n^2+qn\pm1).$$
Equating the coefficients of the $n^3$ terms on each side, we see that $b+q=6$.
Equating the coefficients of the $n^2$ terms on each side, we see that either $bq+2=11$ (if we take the $\pm$ signs to be $+$) or that $bq-2 = 11$ (if the $\pm$ signs are $-$).
The equations $b+q=6$ and $bq-2 = 11$ certainly have no solution in integers, since one of $b$ or $q$ must be 13. So we try the other pair of equations, $b+q=6$ and $bq+2 = 11$. Here the solution $b=q=3$ is obvious, and we are done, except to check the answer. We must do this, because we have completely ignored the coefficient of the $n$ term.