I don't know if this question is still interesting for someone, but in 2008 Argyron and Haydon constructed in "A hereditarily indecomposable L∞-space that solves the scalar-plus-compact problem. Acta Math. 206 (2011), no. 1, 1–54" a infinite-dimensional Banach space in which every operator is the sum of a compact operator and a multiple of the identity. Since the spectrum of compact operators consist of a sequence of point that accumulates only at zero and a escalar perturbation only translate this spectrum, I think this kind of space is pretty suitable for your cuestion. Just take a connected compact set $K$ that is not a singleton and you cannot have operators $T$ with $\sigma(T) = K$. Moreover, it is enough your compact set to have two or more accumulation points to have this property.
It is interesting to mention that one of the other "applications" these space has is to solve a conjectute about the invariant subspace problem. By Lomonosov theorem every compact operator has a non-trivial hyperinvariant subspace, so every escalar perturbation of a compact operator also has a hyperinvariant subspace too. This is the first non-trivial example of an infinite-dimensional space such that every operator has a non-trivial hyperinvariant susbpace.
In
Argyros, Spiros A.; Arvanitakis, Alexander D.; Tolias, Andreas G.
Saturated extensions, the attractors method and hereditarily James tree spaces.
Methods in Banach space theory, 1–90, London Math. Soc. Lecture Note Ser., 337, Cambridge Univ. Press, Cambridge, 2006
the authors construct a separable space $Z$ so that $X:= Z^*$ is non separable, hereditarily indecomposable (HI), and every strictly singular operator on $X$ is weakly compact and hence has separable range. Of course, $X$ embeds isometrically into $\ell_\infty$ because $Z$, being separable, is a quotient of $\ell_1$. There is no operator from this $X$ into $X \oplus X$ that has dense range. Indeed, if $T$ is such an operator, then letting $P_i$ for $i= 1,2$ be the projections onto the factors of $X \oplus X$, we see that $P_iT$ is of the form $\lambda_i I + W_i$ with $W_i$ strictly singular by the usual Gowers-Maurey HI theory. Both operators $P_iT$ have dense range, so neither $\lambda_i$ is zero because both $W_i$ have separable range.
To see that $T$ cannot have dense range, observe that $T^*$ is not injective. Indeed, take a norm one $x^*$ in $X^*$ which vanishes on $W_1X+W_2X$. Then $T^* (\lambda_1^{-1}x^*, - \lambda_2^{-1}x^*)= 0$.
There is also no injective dense range operator from $X\oplus X$ into $X$. For suppose that $S(x_1,x_2) = S_1 x_1 + S_2 x_2$ is such an operator. Write $S_i= \lambda_i I + W_i$ with $W_i$ strictly singular. Since the $W_i$ have separable range and $S$ has dense range, one of the $\lambda_i$, say $\lambda_1$, is not zero, and hence $S_1$ is Fredholm of index zero. If $S_2$ has finite rank, then clearly $S$ is not injective, but otherwise the infinite dimensional $S_2X$ intersects nontrivially the finite codimensional $S_1X$, which again implies that $S$ is not injective.
Best Answer
According to the recent preprint by Koszmider, Shelah and Świętek under the generalised continuum hypothesis there is no such bound. In particular, one cannot prove the existence of such a bound working merely within the ZFC.