Your statement of the Pitman-Koopman-Darmois theorem is off; there is an additional assumption that the support of $\mathcal X$ does not change as $\theta$ changes where $\mathcal X$ is the support of $X_1$ and $\theta$ parameterizes the family. A quick counterexample to the statement of the theorem as given in the OP is the family $\{\mbox{Uniform}(0, \theta): \theta > 0\}$ for which $\max\{X_j, 1 \le j \le n\}$ is sufficient and does not vary with the sample size $n$.
More in the spirit of your question, the answer is yes, there do exist distributions who sufficient statistics are "lossy" even when the conditions of the PKD theorem are satisfied. Consider $X_1, X_2, ...$ iid from a Gamma distribution with shape parameter $\alpha$ (known) and mean $\mu$, and $Z_1, Z_2, ...$ iid Bernoulli with success probability $p$ also known. Then take $Y_i = Z_i X_i - (1 - Z_i)$, and our sample becomes $Y_1, Y_2, ...$. We only get information about $\mu$ when $Y_i \ne -1$, so our sufficient statistic is $\sum_{i: Y_i \ne -1} Y_i$ which grows like $pn$ on average.
Sublinear growth is possible, proceeding along the train of thought suggested above, i.e. using mixtures of distribution, and indeed this is getting at something that is useful in practice. Take $(X_1, Z_1), (X_2, Z_2), ...$ to be iid distributed according to an infinite mixture of normals $f(x, z | \pi, \mu) = \prod_{i = k} ^ \infty [\pi_k N(x | \mu_k, 1)]^{I(Z_i = k)}$, with $Z_i$ an indicator of which cluster $X_i$ is in (I'm not sure if the representation of it via a density that I wrote is valid but you should get the general idea); the dimension of the sufficient statistics should increase only when new clusters are discovered, and the rate of the appearance of new clusters can be controlled by taking $\{\pi_k\}_{k = 1} ^ \infty$ to be known and carefully choosing them; my hunch is that it should be easy to make it grow at a rate of $\log(1 + Cn)$ since I think this is how fast the number of clusters grows in the Dirichlet process.
There is not suitable to write density in the form of a sum. Rewrite it as a product:
$$
f(X=y_1|p_1,\cdots,p_n)=p_1^{{\bf 1}_{x_1}(y_1)}\cdots p_n^{{\bf 1}_{x_n}(y_1)}.
$$
And then
\begin{eqnarray}
f(y_1,\cdots,y_N|p_1,\cdots,p_n)&=&\prod_{i=1}^Nf(y_i|p_1,\cdots,p_n)\\
&=&\prod_{i=1}^N\prod_{j=1}^n p_j^{{\bf{1}}_{x_j}(y_i)} \\
&=& \prod_{j=1}^n p_j^{\sum_{i=1}^N{\bf{1}}_{x_j}(y_i)}. \tag{2}
\end{eqnarray}
This is a function that depends on $p_1,\ldots,p_n$ and on a sufficient statistics
$$
T(y_1,\ldots,y_n) = \left(\sum_{i=1}^N{\bf{1}}_{x_1}(y_i),\ldots,\sum_{i=1}^N{\bf{1}}_{x_{n-1}}(y_i)\right).
$$
Note that really there are $n-1$ unknown parameters since $p_n=1-p_1-\ldots-p_{n-1}$. Also $T$ is $(n-1)$-dimensional since the missed last sum of indicators can be found from the others:
$$
\sum_{i=1}^N{\bf{1}}_{x_n}(y_i) = N-\sum_{j=1}^{n-1}\sum_{i=1}^N{\bf{1}}_{x_j}(y_i).
$$
The expression (2) can be rewritten as
$$
f(y_1,\cdots,y_N|p_1,\cdots,p_n) = \prod_{j=1}^{n-1} p_j^{\sum_{i=1}^N{\bf{1}}_{x_j}(y_i)}\times (1-p_1-\ldots-p_{n-1})^{N-\sum_{j=1}^{n-1}\sum_{i=1}^N{\bf{1}}_{x_j}(y_i)}
$$
$$
=g(T,p_1,\ldots,p_{n-1})\cdot \underbrace{h(y_1,\ldots,y_n)}_1.
$$
Best Answer
For the classical Gaussian distribution, if you have a single observation then your sufficient statistics is $(x,x^2)$ but if you have multiple i.i.d. observations, then the sufficient statistics is $(\sum_{i=1}^n x_i, \sum_{i=1}^n x_i^2)$, and it very well depends on $n$.
You can say that the dimension of sufficient statistics is not increasing with $n$ but the sufficient statistics itself still depends on $n$.