The continued fraction length is usually a small constant factor away from the regulator. A more precise version can also be achieved, but I don't remember a reference, so if anyone does...
Then, we know the regulator times the class number is usually a small constant factor away from the discriminant (of the order, not necessarily the field).
In addition, if a discriminant has $n$ prime factors in its squarefree part, the class number will be divisible by $2^{n-1}$.
Finally, most positive numbers of some size don't have many prime factors, and we suspect the real quadratic fields composed from these to have relatively small class number (look up Cohen-Lenstra).
Combining these facts and heuristics we get that primorials, and even factorials (still have large squarefree part), will have larger class number, hence small regulator, and therefore smaller continued fraction period length.
That said, we can dig even deeper. For factorials, there is a hefty squareful part. When we go from the maximal order of the field $\mathbb{Q}(\sqrt{n!})$ to the order $\mathbb{Z}(\sqrt{n!})$, the discriminant is enlarged hugely. Each prime (even) power $p^{2m}\ ||\ n!$ adds $p^{m-1}(p-(squarefree(n!)/p))$ to the original $h_K\times R_K$. So for each such factor, something goes into the class number of the smaller order, and something goes into the regulator of the smaller order.
Here comes the interesting bit. The factors that make up the new regulator tell us how far the unit group in the small order is from the unit group of the maximal order. Since we are in a real quadratic field the unit groups are of rank 1, so this distance is just the exponent to which the fundamental unit is powered by in order to enter the smaller order. Say $p_1-1$ and $p_2-1$ (say Legendre symbol is 1) have a large gcd. Once we power the fundamental unit to, say, $p_1-1$ to get (locally) into the $p$-part of the smaller order, we can slack off when getting into the $p_2$-part of the smaller order, because we've already done some of the work.
So in the factorial case, or in any number with many square factors (not dividing the squarefree part), since many of the above mentioned factors will have a large gcd, most of the factors will have to go into the class number of the small order - hence the very small regulator and continued fraction expansion.
We can write the above explicitly, but we need another notion. When a prime divides the discriminant to an even power greater than 2, the power of $p$ from the above mentioned factor that goes into the regulator measures how the fundamental unit is far p-adically from being 1. For most $p$, the fundamental unit will not be $1\ (mod\ p^2)$, so pretty much all of the $p^{m-1}$ goes into the regulator.
Hence, we expect:
$$ \frac{period-length(n!)}{\sqrt{n!}} \sim \frac{lcm(\{\ p-(p/squarefree(n!))\ |\ p^{2m}\ ||\ n!\ \}}{\prod_{p^{2m}||n!} (p-(p/squarefree(n!))} $$
And how that factor behaves - I have no idea. Sounds like a combinatorical answer could exist.
Claude Levesque, On semi-reduced quadratic forms, continued fractions and class number, quotes a theorem of Lu, H., On the class number of real quadratic fields, Scientia Sinica II (special number, 1979), 118-130, as follows:
Let $m>1$ be a squarefree integer. Then the class number of ${\bf Q}(\sqrt m)$ is one if and only if $$\theta+\sum_{i=1}^{\ell}k_i=\lambda_1(m)+\lambda_2(m)$$ where $\omega=(1+\sqrt m)/2$ if $m\equiv1\bmod4$, otherwise $\omega=\sqrt m$, the continued fraction for $\omega$ is $[k_0,\overline{k_1,\dots,k_{\ell}}]$, $\theta$ is zero, one, or two depending on $m\bmod4$ and the parity of $\ell$ and $k_{\ell/2}$, and $\lambda_1(m)$ (respectively, $\lambda_2(m)$) is the number of solutions in nonnegative integers of $x^2+4yz=\Delta$ (respectively, $x^2+4y^2=\Delta$), where $\Delta$ is $m$ if $m\equiv1\bmod4$, otherwise $4m$.
For the detailed definition of $\theta$, see the paper of Levesque. The paper goes on to prove related results.
Another paper that may be relevant is Louboutin, Mollin, & Williams, Class Numbers of Real Quadratic Fields, Continued Fractions, Reduced Ideals, Prime-Producing Quadratic Polynomials and Quadratic Residue Covers, Canadian Journal of Mathematics 44 (1992) 824-842. DOI:10.4153/CJM-1992-049-0.
Best Answer
As you indicate, real algebraic numbers of degree $\leq 2$ have this property in view of Lagrange's classical result characterizing them by the eventual periodicicty of the continued fractions expansion. It may be useful to know (if you don't already) that $\alpha \in \mathbb{R}$ having bounded continued fractions coefficients is equivalent to the sharpness of Dirichlet's approximation theorem: $|\alpha - p/q| > cq^{-2}$ for all but finitely many $p/q \in \mathbb{Q}$.
For algebraic numbers this means that the exponent $2+\varepsilon$ in Roth's theorem can be reduced to $2$. For quadratic irrationalities this holds with the uniform constant $c = 1/\sqrt{5} - \epsilon$; google "Lagrange spectrum." As far as I know it is widely believed that this may not happen for any algebraic number of degree $> 2$, although Serge Lang has suggested that a milder improvement in Roth's theorem, whereby $q^{2+\varepsilon}$ gets replaced with $(q\log{q})^2$, is always possible. (This is wide open; there is an analogous statement known to hold true in Nevanlinna theory). Certainly no algebraic number of degree $> 2$ is known to have unbounded partial fractions coefficients, but worse, it does not appear to be even known that there exists an algebraic number having this property.
A reference where this appears explicitly in print is p. 366 of Hindry and Silverman's Diophantine Geometry: An Introduction.
Added. There is also a rather interesting variant of this question for $\mathbb{Z}[i]$-continued fractions expansions of complex algebraic numbers.
As is typical in diophantine analysis, both Roth's theorem and the continued fractions algorithm extend to the relative setting over number fields other than $\mathbb{Q}$; and to a large extent, so does the relationship between the two. To be concrete, consider rational approximations over the Gaussian field $\mathbb{Q}(i)$. The relative Roth theorem over $\mathbb{Q}(i)$ states: If $\alpha \in \mathbb{C}$ is algebraic, then for every $\varepsilon > 0$ there are only finitely many pairs $p,q \in \mathbb{Z}[i]$ in the Gaussian lattice satisfying $|\alpha - p/q| < |q|^{-2-\varepsilon}$. [For the general statement over any number field see Thm. 6.2.3 of Bombieri and Gubler's Heights in Diophantine Geometry.]
Likewise, Hurwitz has attached to a complex number a canonical continued fractions expansion with entries in $\mathbb{Z}[i]$, by using the same algorithm as over $\mathbb{Q}$, but with the nearest rounding to the Gaussian lattice $\mathbb{Z}[i]$ (whereas over $\mathbb{Q}$ the conventional choice of rounding involves the floor function $\lfloor \cdot \rfloor$ instead of the nearest rounding to $\mathbb{Z}$). The trichotomy "rational-quadratic-higher degree" would appear to extend to the relative setting over this (or any other) number field $\mathbb{Q}(i)$: the $\alpha \in \mathbb{C}$ with terminating expansion are of course precisely the numbers in $\mathbb{Q}(i)$, and Hurwitz has shown that his expansion is eventually periodic if and only if $[\mathbb{Q}(\alpha,i):\mathbb{Q}(i)] \leq 2$.
We can ask, then, the same questions for Hurwitz's complex continued fractions. Rather suprisingly, there are algebraic numbers whose $\mathbb{Z}[i]$-continued fractions expansion has bounded coefficients, and whose relative degree over $\mathbb{Q}(i)$ is $> 2$! One such number, due to D. Hensley [1, Ch. 5] in 2006, is $\sqrt{2} + i\sqrt{5}$, of relative degree four. More generally, W. Bosma and D. Gruenewald [3] have shown that a complex number has this property if the square of its modulus is a rational integer which is not a norm from $\mathbb{Z}[i]$; algebraic such examples thus include $\sqrt[m]{2} + i\sqrt{n - \sqrt[m]{4}}$ for all $n \equiv 3 \mod{4}$ and $m$.
(On the other hand, to my knowledge no particular algebraic number has been proven to have unbounded coefficients in its Hurwitz $\mathbb{Z}[i]$-continued fractions expansion.)
But now there is something even more curious about the implication of such examples for Roth's theorem over $\mathbb{Q}(i)$ (diophantine approximations by Gaussian numbers). I realize this just now after looking up Hensley's very interesting paper [2] (from 2006), and I wonder why this point isn't brought up in the literature on complex continued fractions.
While the convergents $p_n/q_n \in \mathbb{Q}(i)$ in Hurwitz's expansion no longer exhaust all good $\mathbb{Q}(i)$-rational approximations to $\alpha \in \mathbb{C}$ (as they do in the case over $\mathbb{Q}$), Theorem 1 in [2] shows that up to a multiplicative constant, they still give the best $\mathbb{Q}(i)$-rational approximations. As a result, for those numbers (algebraic of arbitrarily high relative degree over $\mathbb{Q}(i)$), the exponent $2+\varepsilon$ in Roth's theorem rel. $\mathbb{Q}(i)$ (stated above) can be reduced to $2$, and this is moreover effective:
Consequence: If $\alpha \in \mathbb{C}$ has $|\alpha|^2 \in \mathbb{Q}$ a rational number not a norm from $\mathbb{Q}(i)$ [e.g., the rel. degree-four algebraic example $\sqrt{2} + i\sqrt{5}$], then there is an effective $c(\alpha) > 0$ such that $|\alpha - \beta| > cH(\beta)^{-2}$ for all $\beta \in \mathbb{Q}(i)$. (Here, $H(\cdot)$ is the absolute multiplicative height on $\bar{\mathbb{Q}}$; for rational or imaginary quadratic integers $n$, it coincides with $|n|$)
Consider now Khintchine's principle according to which an algebraic number of degree $> 2$ should be generic. As almost every complex number $x \in \mathbb{C}$ has infinitely many $\mathbb{Q}(i)$-rational approximants $\beta \in \mathbb{Q}(i)$ with $|x - \beta| < 1/(H(\beta)^2\log{H(\beta)})$, the numbers $\alpha$ of the above form are not generic in this sense. As they contain algebraic numbers of arbitrarily high (though necessarily even) rel. degree over $\mathbb{Q}(i)$, Khintchine's principle would appear to fail in the relative setting over $\mathbb{Q}(i)$!
Nevertheless I think that the story is more interesting, as the above examples still resemble quadratic irrationalities in some vague sense. Perhaps, in the relative setting of diophantine approximations over a number field $K$, we could salvage Khintchine's principle and the above trichotomy by enlarging the class of special algebraic numbers -- which a priori contain all numbers of rel. degree $\leq 2$ over $K$.
Allow me to record a few
Problems. Might the numbers of degree $\leq 2$ over $\mathbb{Q}(i)$ (whose Hurwitz expansions are eventually periodic) and the algebraic numbers mentioned in the above "Consequence" (whose Hurwitz expansions are aperiodic yet have bounded coefficients) exhaust all the algebraic numbers for which the exponent $2+\varepsilon$ in Roth's theorem rel $\mathbb{Q}(i)$ may be reduced to $2$? Should we expect that all algebraic numbers not of this shape ought to satisfy Khintchine's principle rel $\mathbb{Q}(i)$? What are the special algebraic numbers in diophantine approximations rel a general given number field $K$? Finally, the same problem can be considered about $p$-adic (and $S$-adic) $K$-rational approximations to algebraic numbers; I do not know if this has been done even for $K = \mathbb{Q}$.
References:
[1] D. Hensley: Continued Fractions (World Scientific, Singapore, 2006).
[2] D. Hensley: The Hurwitz complex continued fraction (2006): http://mosaic.math.tamu.edu/~dhensley/SanAntonioShort.pdf
[3] W. Bosma, D. Gruenewald: Complex numbers with bounded partial quotients, J. Aust. Math. Soc., vol. 93 (2012), pp. 9--20.