Suppose I have a norm which is composed of two norms in a convex fashion:
$$||x|| = \alpha ||x||_1 + (1-\alpha)||x||_2,$$ where $\alpha \in [0,1]$. Now suppose I'm interested in finding the subdifferential of $||x||$ at zero, i.e, the unit ball of the dual norm. Can I derive the unit ball of $||x\|$ in terms of those for $||x||_1$ and $||x||_2$? These are not the traditional $\ell_1$ and $\ell_2$ norms, but generic norms. Formally, I'm looking to derive the set
$$B_{||\cdot||^*}[0,1] = \{y: ||y||^* \leq 1\}$$
in terms of $B_{||\cdot||_1^*}[0,1]$ and $B_{||\cdot||_2^*}[0,1]$.
The new norm you've described makes your new space sit isometrically as a subspace of a $\oplus_1$ sum, so the dual will be a quotient of an $\oplus_\infty$ sum. The quotient involves taking an infimum over decompositions, so the result is potentially intractable.
Since $\|\cdot\|$ is equal to either $\|\cdot\|_1$ or $\|\cdot\|_2$ if $\alpha\in \{0,1\}$, I will assume $0<\alpha<1$.
POINT $1$ We know that if $E$ is a subspace of $F$, then $E^*=F^*/E^\perp$, where $E^\perp=\{f^*\in F^*:f^*|_E\equiv 0\}$. That is, duals of subspaces are quotients (and duals of quotients are subspaces, but we won't use that here).
More generally, if $j:G\to F$ is an isometric embedding (a linear operator with $\|j(g)\|_F=\|g\|_G$ for all $g\in G$), then $G^*\approx F^*/j(G)^\perp=F^*/\ker(j^*)$, where $\approx$ denotes isometric isomorphism. That's because $G$ is isometrically isomorphic to $E:=j(G)$ (via $j$, of course), so $G^*$ is isometrically isomorphic to $E^*=F^*/j(G)^\perp$, and $j(G)^\perp=\ker(j^*)$ by usual nonsense. In fact, $j^*$ can be thought of as the quotient map from $F^*$ to $F^*/\ker(j^*)\approx G^*$, in the sense that $$\|g^*\|_{G^*}=\inf\{\|f^*\|_{F^*}:j^*f^*=g^*\}.$$
POINT $2$ If $(E,e)$ and $(F,f)$ are Banach spaces, then for $1\leqslant p\leqslant \infty$ and $q$ with $1/p+1/q=1$, $$((E,e)\oplus_p (F,f))^*=(E^*,e^*)\oplus_q (F^*,f^*).$$
Suppose we have $\alpha\in [0,1]$ and two norms $\|\cdot\|_1$, $\|\cdot\|_2$ on the same vector space $X$ such that $(X, \|\cdot\|_1)$, $(X, \|\cdot\|_2)$ are Banach spaces. To make things prettier, let $X_1$ denote $X$ endowed with the norm $\|\cdot\|_1$ and let $X_2$ denote $X$ endowed with the norm $\|\cdot\|_2$. Similarly, let $X_1^*$, $X^*_2$ denote $X^*$ endowed with $\|\cdot\|^*_1$, $\|\cdot\|_2^*$, respectively. Define $j:X\to X_1\oplus_1 X_2$ by $$j(x)=(\alpha x, (1-\alpha)x).$$ Define a new norm $\|\cdot\|$ on $X$ by $$\|x\|=\|j(x)\|=\alpha \|x\|_1+(1-\alpha)\|x\|_2.$$ So $j:(X, \|\cdot\|)\to X_1\oplus_1 X_2$ is an isometry (we forced this with the definition of $\|\cdot\|$). Combining the first two points, $(X^*, \|\cdot\|^*)$ is isometrically identifiable with $$(X_1\oplus_1 X_2)^*/\ker(j^*) = (X^*_1\oplus_\infty X^*_2)/\ker(j^*).$$
Note that $j^*:X^*_1\oplus_\infty X^*_2\to X^*$ is given by $$j^*(x^*_1,x^*_2)=\alpha x^*_1+(1-\alpha)x^*_2.$$ Indeed, \begin{align*}(\alpha x^*_1+(1-\alpha)x^*_2)(x)=x^*_1(\alpha x)+x^*_2((1-\alpha)x)=(x^*_1,x^*_2)j(x). \end{align*}
So for any $x^*\in X^*$, we have from point $1$ that $$\|x^*\|^* = \inf\{\max\{\|x^*_1\|^*, \|x^*_2\|^*_2\}:x^*_1,x^*_2\in X^*,\alpha x^*_1+(1-\alpha)x^*_2=x^*\}.$$ This can be rewritten as $$\|x^*\|^* = \inf\Bigl\{\max\bigl\{\bigl\|\frac{x^*-y^*}{\alpha}+y^*\bigr\|_1^*,\|y^*\|_2^*\bigr\}: y^*\in X^*\Bigr\},$$ although I'm not sure if that's any more tractable than the preceding expression.
Two other equivalent expressions would be $$\|x^*\|^* = \inf\Bigl\{\max\{\alpha^{-1}\|x^*_1\|^*_1, (1-\alpha)^{-1}\|x^*_2\|^*_2\}:x^*_1,x^*_2\in X^*, x^*_1+x^*_2=x^*\Bigr\},$$ or $$\|x^*\|^* = \inf\Bigl\{\max\{\alpha^{-1}\|y^*\|^*_1, (1-\alpha)^{-1}\|x^*-y^*\|^*_2\}:y^*\in X^*\Bigr\}.$$