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.