I know that this post is quite old. However, the answer is maybe interesting for you. Directly applying lensmaker's equation doesn't work due to the air inside the lens.
I would rather suggest the following:
Since the middle of such a shell consists of air we can conceptually cut it in two halves.
These two halves are two identical lenses (just turned by 180°).
The focal length of one half is:
$$ \frac{1}{f} = (n-1) \left(\frac1{R_1} - \frac1{R_2} + \frac{(n-1) d_\text{lens}}{n R_1 R_2}\right)$$
Note that both $R_1$ and $R_2$ are positive. In our special case $R_2 < R_1$ and so this lens acts as a diverging lens. $d_\text{lens} = R_1 -R_2$ is the thickness of our lens.
Now we've got a lens system of two such lenses and we have to combine those. This done via:
$$\frac{1}{f_{res}} = \frac1{f_1} + \frac1{f_2} - \frac{d}{f_1 f_2}$$
where $d$ is the distance between this lenses (two spherical shell halves). In our case it is $d=2\cdot \frac{R_1+R_2}{2}$.
To finally answer your question. In the case of a zero meniscus lens we get $R_1 = R_2$ and also $d_\text{lens} = 0$. This means that the focal length is 0. So you won't see any effect of the spherical shell.
edit: this whole calculation neglects spherical aberrations. So this is only valid for rays close to the center of the shell.
