Will cinnamaldehyde undergo self-aldol condensation?
Not every &#x3b1;-hydrogen automatically means that a ketone is enoliseable. It would need to be an &#x3b1;-hydrogen capable of aligning with the &#x3c0; system of the $\ce{C=O}$ bond such that removing the hydrogen automatically generates the required enolate anion system. In the case of cinnamaldehyde — your compound — there is conjugation across the phenyl ring, the double bond and the aldehyde meaning that this compound is entirely planar. Thus, there is no way for the &#x3b1;-hydrogen to align favourably for deprotonation to occur. If it did, it would generate an allene-olate $\ce{Ph-CH=C=CH-O-}$ which is not particularly favourable.

In general, &#x3b1;,&#x3b2;-unsaturated ketones and aldehydes are deprotonated vinylogously, meaning that the double bond shifts away upon deprotonation. For example crotonaldehyde $\ce{H3C-CH=CH-CHO}$ is deprotonated in the following way:

$$\ce{H3C-CH=CH-CHO -&gt;[][- H+] H2C=CH-CH=CH-O-}\tag{1}$$

I.e. the &#x3b3;-hydrogen is abstracted in place of the &#x3b1;-hydrogen.