Why would a pendulum experiment give $g > 9.8\ \mathrm{m/s^2}$? I am taking an introductory lab course in which we've done an experiment on the physical pendulum.

We've seen that for small oscillations, the period is 
$$T=2\pi\sqrt{\dfrac{I_S}{Mgd_{cm}}}\tag{1}$$
where $S$ is the pivot point, $M$ is the total mass of the object, and $d_{cm}$ is the distance between $S$ and the pendulum's center of mass. 
Now, varying $d_{cm}$, I've obtained seven different periods. I've calculated $g$ in terms of $T$ and $d_{cm}$ for those seven distinct values of $T$ and $d_{cm}$. So, $g$ can be expressed as 
$$g=4\pi^2\dfrac{I_S}{T^2Md_{cm}}.\tag{2}$$ 
All the periods I've obtained were always greater than $1$, and the values of $g$ where between $10.15$ and $10.3$. I am trying to understand why is it that $g$ gave me always greater than $9.8$, the expected value, and not less than $9.8$. If I consider the air friction, I would expect the period to be greater; from what I've said and from equation (2), I would say that the values of $g$ should be less than $9.8$, contrary to the values I got.
Note that that $T$ is in seconds, $[g]=\dfrac{m}{s^2}$ and the angle from which the pendulum was released is of $25$ degrees approximately.
I would appreciate if someone could help me to understand the reason why I've obtained these values for $g$.
 A: I) OP is using the period formula 
$$\tag{1} T~=~2\pi\sqrt{\frac{I}{MgR}} $$
for a compound/physical pendulum (in the small amplitude limit) to estimate the gravitational acceleration constant
$$\tag{2} g~=~\left(\frac{2\pi}{T}\right)^2 \frac{I}{MR}. $$
Here $I$ is the moment of inertia around the pivot point; $R$ is the distance from CM to the pivot point; and $M$ is the total mass. 
II) After doing the experiment OP finds values for $g$ that are 3-5% too big. (These results are close enough that OP likely did not make any elementary mistakes with units.) A finite amplitude of 
$$\tag{3} \theta_0 ~\approx ~25^{\circ}~\approx~ .44~ {\rm rad}$$ 
makes the pendulum 
$$\tag{4} \frac{\theta_0^2}{8}~\approx~ 2\%$$ 
slower, as compared to the ideal pendulum (1), cf. comment by Prahar. So correcting for a finite amplitude makes OP's estimates worse, 5-7% too big, as Keith Thompson points out in a comment above.
So the discrepancy is caused by something else. The culprit is likely that it is difficult to get a precise estimate for the moment of inertia $I$. All the other quantities $T$, $M$ and $R$ should be fairly easy to measure reliable. So OP's value for $I$ is likely too big. According to Steiner's theorem 
$$\tag{5} I~=~MR^2+I_0,$$ 
where $I_0$ is the moment of inertia around the CM (and the actual quantity which is poorly known).
III) Below follows a suggestion. Plot OP's seven data points in an $(x,y)$ diagram with axes
$$\tag{6} x~:=~R^2 \quad\text{and}\quad y~:=~R\left(\frac{T}{2\pi}\right)^2.$$ 
Theoretically, the $(x,y)$ data points should then lie on a straight line 
$$\tag{7} y~=~ax+b$$
with slope 
$$\tag{8} a~=~\frac{1}{g}$$ 
and $y$-intercept 
$$\tag{9} b=~~\frac{I_0}{gM}.$$ 
In other words, find the best fitting straight line. This method should hopefully produce a good estimate for $g$ without having to know $I_0$ a priori. (By the way, notice that we in principle also don't need to know the mass $M$, cf. the equivalence principle!)
IV) Finally, as always in experiments, estimate all pertinent uncertainties in the various measurements.
