About the existence of characters on $B(X)$

I guess that you mean that $B(H)$ has no character (=continuous unital algebra homomorphism into $\mathbf{C}$) if $H$ has dimension $\neq 1$ (idem for $M_n(\mathbf{C})$ for $n\neq 1$), and thus that your question assumes $\dim(X)\ge 2$ (and hence $=\infty$).

Argyros and Haydon (Acta Math, 2011: arXiv, Project Euclid unrestricted access) constructed a Banach space $X$ of infinite dimension in which every bounded self-operator is scalar+compact. Hence for such a space, modding out by the ideal of compact self-operators yields a character.

Examples were known before the Argyros-Haydon space mentioned in Yves Cornulier's answer. For instance, if $J$ denotes the James space, then the image of the canonical map $J\to J^{**}$ has codimension $1$, and from this one can show that the closed 2-sided ideal $W(J)$ of all weakly compact operators on $J$ has codimension one in $B(J)$, thus giving us a character.

I am not sure where this was first observed: I learned of it from this paper of Loy and Willis:

MR1044280 (91f:46069) R. J. Loy, G. A Willis. Continuity of derivations on B(E) for certain Banach spaces E. J. London Math. Soc. (2) 40 (1989), no. 2, 327–346.

In the same paper, they exhibit some other examples of $E$ for which $B(E)$ admits a closed ideal of codimension $1$. For instance, one can take $K$ to be the scattered compact space obtained from the interval $[0,\omega_1]$ in the order topology, and then $E=C(K)$ has this property. (An alternative construction of the character for $B(C(K))$ can also be found in the 2014 paper of Kania, Koszmider and Lauststen, "A weak∗-topological dichotomy with applications in operator theory".) They also construct a James-type example of an $E$ such that $B(E)$ quotients on $\ell^\infty$, and hence has a lot of characters. All these examples are of a different flavour to the spaces that have been constructed more recently using the Argyros-Haydon machinery and its descendants.

At the other extreme, it is worth noting that if $X=\ell_p$ ($1\leq p\leq\infty$) or $X=L_p[0,1]$ ($1\leq p \leq\infty$) then $B(X)$ has no characters.

In order to complement answers given by Ycor and Yemon Choi, let me mention the space $X_M$ constructed by Mankiewicz (Isreael J. Math., 1989), which has the following remarkable properties:

• $X_M$ is separable and super-reflexive,
• $B(X_M)$ has a continuous homomorphism onto $\ell_\infty$.

Since $\ell_\infty$ has $2^{2^{\aleph_0}}$ characters, so has $B(X_M)$.

Another example of a space $X$ for which $B(X)$ has a character is G. Edgar's long James space. Recently, many spectacular separable examples of spaces whose algebras of operators admit characters have been constructed so the list goes on.