Negative. This fails already for large enough representations with $k=2$. Barring a programming error the first counterexample I found occurs with the 5-dimensional representation.
I don't want to carry out computations like this by paper & pencil alone, so I needed to reformulate this with something I can plug into a CAS (in my case Mathematica).
The generators of this Lie algebra (= the Pauli spin matrices) act on the 2-dimensional space $V$ spanned by two vectors $u$ and $v$ that I choose to treat as variables via the partial differential operators:
$$
\begin{aligned}
S_x(p)&=v\frac{\partial p}{\partial u}+u\frac{\partial p}{\partial v},\\
S_y(p)&=i v\frac{\partial p}{\partial u}-i u\frac{\partial p}{\partial v},\\
S_z(p)&=u\frac{\partial p}{\partial u}-v\frac{\partial p}{\partial v}.\\
\end{aligned}
$$
My motivation for using partial differential operators is twofold. Most importantly the exact same operator relation then also applies to the symmetric powers $Sym^n(V)$, this is well known (at least in the case of the complexification $\mathfrak{sl}_2$ that I am more familiar with).
The symmetric power $Sym^n(V)$ is exactly the $(n+1)$-dimensional irreducible representation of $\mathfrak{su}_2$, so it suffices to scan the list of them to settle this question. Also, $Sym^n(V)$ is then the space of homogeneous polynomials of degree $n$ in the two unknowns $u,v$. This takes us to my second motivation — it is easy to program a CAS to apply an algebraic combination of partial differential operators to a generic homogeneous polynomial!
Concentrating on the case $k=2$, so the operator
$$
\Delta:=S_x^4+S_y^4+S_z^4.
$$
It turns out (make your computer eat the program) that, for example,
$\Delta(p)=123 p$ for all homogeneous cubic polynomials $p(u,v)$. I believe that examples like this guided the OP to their conjecture.
However, in $Sym^4(V)$ see the following:
$$
\begin{aligned}
\Delta v^4&=48(u^4+7v^4),\\
\Delta v^3u&=288 v^3u,\\
\Delta v^2u^2&=384 v^2u^2,\\
\Delta vu^3&=288 vu^3,\\
\Delta u^4&=48(v^4+7u^4).
\end{aligned}
$$
So $\Delta$ does not act as a scalar on the 5-dimensional irreducible representation. Consequently it won't belong to the center of the universal enveloping algebra either.
It may be worth noting that the operator
$$
L^2:=S_x^2+S_y^2+S_z^2
$$
does act as the scalar $n(2+n)$ on the irreducible representation $Sym^nV$. This reflects the fact (familiar to all students of quantum mechanics under a different guise in relation to spin/angular momentum) that $L^2$, indeed, is in the center of $U(\mathfrak{su}_2)$.
