Here is a proof which works for a large class of measures $\mu$, including the example $\mu(n)=(n+1)^2$ of the OP. I only consider measures whose support is $\mathbb{N}=\{0, 1, \ldots\}$.

As noted by @JeanDuchon, it suffices to show that $H(X_n) = \Omega(\log n)$ (with the Big-Omega notation).

*The key formula:* For any random variable $X$ supported by $\{0, \ldots, n\}$, it can be shown that
$$
\boxed{H(X) = \mathbb{E}\left[\dfrac{h(p_X)}{p_X}\right]}
$$
where $h(p)=-p\log p -(1-p)\log(1-p)$ and $\boxed{p_k=\dfrac{\Pr(X=k)}{\Pr(X \leq k)}}$ ($(p_k)$ is called the *reversed hazard rate* in some papers).

I discovered this elementary but useful formula a couple of years ago. Some details are currently given on my blog (in French) and I will soon deposit a preprint or a working paper about that. I would find surprising that nobody discovered this formula before me, but oddly, I have not been able to find it in the literature.

As a consequence, given a measure $\mu$ supported by $\mathbb{N}$, the entropy of a random variable $X_n \sim \mu( \cdot \mid 0:n)$ can be written
$$
H(X_n) = \mathbb{E}\left[\frac{h(p_{X_n})}{p_{X_n}}\right]
$$
where $\boxed{p_k=\dfrac{\mu(k)}{\mu(0:k)}}$.

Consider the case $\mu(n) = (n+1)^2$. Then the $p_n$ (derived by @JeanDuchon) are decreasing. Since the function $h(x)/x$ is decreasing,
$$H(X_n) = \mathbb{E}\left[\frac{h(p_{X_n})}{p_{X_n}}\right] > \dfrac{h(p_K)}{p_K}\Pr(X_n >K) > -\log p_K \Pr(X_n >K).$$

Moreover $-\log p_n = \Omega(\log n)$. Taking $K=[n/2]$,
$$H(X_n) > -\frac{1}{2}\log p_{[n/2]} = \Omega(\log n),$$
the desired result.

This proof more generally works for $\mu(n)$ increasing in $n$, having decreasing $p_n$ (or even something weaker) and such that $-\log p_n = \Omega(\log n)$. With my (current) terminology, this implies that such a measure $\mu$ has *full entropy*. Roughly speaking, *that means that for large $n$, you cannot approximate $X_n$ by a random variable having form $f(X_n)$ and having entropy lower than $H(X_n)$*.
I can also derive $H(X_n) \sim \log n$ from these investigations.

As a side note, this claim in my OP:

I know how to prove that $H(X \mid X >n) \leq H(X)$ as well as $H(X \mid X <n) \leq H(X)$ for every $n$

is also shown with the help of the integral represententation $H(X_n) = \mathbb{E}\left[\frac{h(p_{X_n})}{p_{X_n}}\right]$ (it follows from this equality and from the monotonicity of the $p_n$).