**edit:** this answer is garbage (or, rather, answers a question that the asker did not ask).  I leave it here because I think the comments are valuable.  If, later, a real answer appears, then I will delete it.

$3$ is the least possible.  

Ingredient 1: For a one-dimensional complete local ring $R$ with integral closure $\tilde R$, the $k$-dimension of $\tilde{R}/R$ is one less than the multiplicity $e(R)$ of the ring.  This is because $e(R) = \dim_k (\tilde {R}/\mathfrak{m}\tilde{R})$, which is due to Greither in 1982.

Ingredient 2: There is an inequality due to Abhyankar for the multiplicity of a CM local ring: $$e(R) \geq \mu_R(\mathfrak{m}) - \dim R + 1$$ where $\mu$ denotes the minimal number of generators.  

Ingredient 3: It's relatively easy to see that for a Gorenstein local ring that is not a hypersurface (e.g. a one-dimensional non-planar Gorenstein local ring) we can do one better than Abhyankar's bound. Namely, if we had equality in Abhyankar's bound, then $\mathfrak m^2 = \mathbf{x}\mathfrak m$ for some minimal reduction $\mathbf{x}$ of the maximal ideal.  Count lengths in $\bar{R} = R/(\mathbf{x})$, remembering that it has one-dimensional socle since $R$ is Gorenstein, to see that $\dim R = \mu_R(\mathfrak m) -1$.  Therefore we have
$$e(R) \geq \mu_R(\mathfrak{m}) - \dim R + 2$$
for a Gorenstein non-hypersurface $R$.

Applying this formula with $\mu_R(\mathfrak m) \geq 3$ and $\dim R = 1$, we get that the multiplicity is at least $4$, so the degree is at least 3.