Hahn-Banach theorem with convex majorant
If $p$ is convex, then $P(x)=\inf_{t>0}t^{-1}p(tx)$ is sublinear, isn't it? Also, if a linear functional is dominated by $p$, it is also dominated by $P$. Finally, $P\le p$. So there is no non-trivial gain in generality whatsoever unless you start talking about extending non-linear functionals but then you should restate the question accordingly.
Other References Containing the Convex Hahn-Banach
The convex majorant version of HBT (Hahn-Banach Theorem) also appears in
"Methods of Modern Mathematical Physics Vol. 1 Functional Analysis" by Reed and Simon (p.75-76)
"Handbook of Analysis and Its Foundations" by Eric Schechter (statement p.318; proof p.321-322).
"Convex" Hahn-Banach for Complex Vector Spaces
Reed and Simon also have a "convex" majorant HBT for complex vector spaces. The standard HBT for complex spaces has a semi-norm $p$ and a complex-linear functional $f$ with $|f| \leq p$. In the ``convex" majorant HBT for complex spaces, $p$ need not be a semi-norm, but is only required to satisfy the weaker condition \begin{align}\label{1}\tag{1} p(\alpha x + \beta y) \leq |\alpha|p(x) + |\beta|p(y) \quad \text{ when } |\alpha|+|\beta|=1. \end{align}
Reed and Simon deduce the "convex" complex HBT from the convex real HBT in the same way that the standard semi-norm complex HBT is deduced from the standard sublinear real HBT (that is, by exploiting the relationship between a complex-linear functional and its real part). And, of course, the proof of the convex real HBT is basically identical to the proof of the sublinear real HBT.
Alternatively, the "convex" complex HBT can be deduced from the semi-norm complex HBT by changing Fedja's $P(x)$ to $P(x)=\inf_{0 \neq s \in \mathbb{C}} |s|^{-1}p(sx)$. Indeed, if $p$ satisfies \eqref{1}, then $P$ is a semi-norm; if a linear functional is dominated by $p$, it is also dominated by $P$; and $P \leq p$.
Who First Noted the Convex HBT
Since the proofs of the convex and sublinear HBT are so similar and in light of Fedja's argument, I would guess that the convex HBT was known basically as soon as the sublinear HBT was. That being said....
Schechter on page 318 points out that the convex HBT is not mentioned much in the literature but was known at least as early as in
- Nakano, Hidegorô. On an extension theorem. Proc. Japan Acad. 35 (1959), no. 3, 127. doi:10.3792/pja/1195524402. https://projecteuclid.org/euclid.pja/1195524402
See also
- Nakano, Hidegorô. An extension theorem. Proc. Japan Acad. 33 (1957), no. 10, 603--604. doi:10.3792/pja/1195524883. https://projecteuclid.org/euclid.pja/1195524883
According to the last reference, the convex real HBT may also appear in the following two references, but I haven't been able to check.
H. Nakano: Modulared linear spaces, Jour. Fac. Sci. Univ. Tokyo, I, 6, 85 (1951).
H. Nakano: Topology and Linear Topological Spaces, Tokyo (1951).
Just two remarks on the above answer but too long for a comment. Firstly, there is no reason why the above sublinear function should be finite everywhere, indeed it can be the constant function $-\infty$. I don't think that this invalidates the argument but it does make it a tad messier.
Secondly, I think that there is a very good reason for using sublinear functions. There is a perfect duality between such functionals on a vector space and between the algebraically closed, convex, absorbent subsets which contain zero. This supplies a transparent link between the analytic and the geometric versions of the result in question (a fact which has been mentioned several times in the above comments). Of course, the sublevel sets of a a convex function are also convex, but, in contrast to the case of a sublinear functional where these are all dilations of its unit ball, these form an infinite family of such sets, typically of different form. Of course, a convincing example of an interesting application of the version involving domination by a convex function would change the whole ball game.