Why can you not make a compound with a quaternary alpha carbon using malonic ester?
Double alkylations of the $\ce{CH2}$ centre of malonic esters do work!

Have a look at this marvel by R. P. Mariella and R. Raube, published in Org. Synth., 1953, 33, 23




EDIT 1

The use of sodium metal in a properly dried alkanol (in the case of ethanol, refluxing over magnesium ribbon is a good choice) was (and still is) an excellent and cheap method to generate sodium alkoxide in situ:

\[\ce{2Na + 2ROH -> 2NaOR + H2}\]

I don't know if nowadays commerically available sodium ethoxide and/or sodium hydride were that much used back in the days when the procedure was published. Fact is that both these can rot away in humid air when somebody leaves the bottles and cans open for too long, and this happens frequently in university labs.

With sodium metal, you're on the safe side: a freshly cut piece of metal is always good to go. 


EDIT 2

For the example given above, no further direct alkylation at that position is possible, and this is exactly, what ron pointed out in his
comment to the question:


  Perhaps the author was trying to say that by using just the malonic ester synthesis you can't prepare a compound with 3 (potentially different) alkyl groups on the alpha carbon.


He also suggested the viable solution, which I had outlined in another question on this site. Upon acidification and heating, $\beta$-ketoesters and 1,3-diesters undergo decarboxylation, which makes them great synthons to introduce $\ce{-CH2CO-CH3}$ and $\ce{-CH2COOH}$, respectively.

Converting a 1,3-diester to the corresponding carboxylic acid this way means that the new structure now again has an $\alpha$-proton that can be abstracted. The resulting enolate can be alkylated.

In the case of a carboxylic acid, this is usually done in a solvent like tetrahydrofurane or 1,4-dioxane, using two bases. For the deprotonation of the acid, $\ce{NaH}$ is sufficient, while the abstraction of the $\alpha$-proton, a strong base like $\ce{LDA}$ is used. (Exactly this is outlined in the textbook for the alkylation of an ester).