Example of a ternary $C^{\ast}$-ring which is not an operator space A ternary $C^{\ast}$-ring  is a complex Banach space $X$, equipped with a ternary product $[\cdot,\cdot,\cdot]:X^3 \to X$ which is linear in outer variables and conjugate linear in middle variable. Also $X$ is associative i.e.  $$[[a,b,c],d,e]=[a,[d,c,b],e]=[a,b, [c,d,e]].$$  Moreover, $\lVert[a,a,a]\rVert= \lVert a\rVert^3$ and $\lVert[a,b,c]\rVert \leq \lVert a \rVert \lVert b\rVert\lVert c\rVert$.

Does there exist a ternary $C^{\ast}$-ring which is not an operator space?

One obvious class of examples of ternary $C^{\ast}$-rings is the class of  ternary rings of operators but they all are operator spaces.
 A: According to Zettl [1], a ternary ring of operators (TRO) is a
ternary $C^*$-ring which is isomorphic to a closed
subspace $X\subseteq B(H)$, such that $XX^*X\subseteq X$, equipped with the ternary multiplication
$$
  [x,y,z] := xy^*z.
  $$
On the other hand, an anti-TRO is a ternary $C^*$-ring defined as above, except that the multiplication operation is
$$
  [x,y,z] := -xy^*z.
  $$
It is a fundamental result of Zettl [1] that every ternary $C^*$-ring $X$ decomposes uniquely as
$$
  X=X_+\oplus X_-,
  $$
where $X_+$ is a TRO, and $X_-$ is an anti-TRO .
It seems to me that the reading of the question posed by the OP that makes the most sense is by taking the expression
"operator space" to mean a TRO.  In this case the answer is yes, there does exist a ternary $C^*$-ring which is not a
TRO: just take any non-zero anti-TRO.  For an even more concrete example, take $X=M_{n\times m}({\bf C})$, with ternary
multiplication $[x,y,z] := -xy^*z$.
On the other hand, if one takes the expression "operator space" for its face value, Zettl's result implies that every
ternary $C^*$-ring is an operator space in an even more canonical form than suggested by user @YemonChoi: write
$X=X_+\oplus X_-$, embedd $X_+$ in $B(H_+)$, and $X_-$ in $B(H_-)$, whence
$$
  X\subseteq B(H_-\oplus H_+).
  $$
This embeding preserves the operator space structure (norms on matrix algebras) that a TRO canonical possesses.
It is interesting to remark that if you change the (binary) multiplication operation on a $C^*$-algebra by
$$
  x\circ y := -xy,
  $$
then the resulting object is strictly speaking a new C*-algebra, but it is isomorphic to the old one.  The isomorphism
is simply $a\mapsto -a$.
However, if you change the (ternary) multiplication on a ternary $C^*$-ring by inserting a minus sign as above, then the
map $a\mapsto -a$ is no longer an isomorphism,  essentially because 2 is even and 3 is odd!
Indeed, Zettl's uniqueness result tells you that the new ternary $C^*$-ring
might not be isomorphic to the old one at all!

EDIT: Here are some details of Zettl's proof which might shed some light into the reason an anti-TRO not isomorphic to a TRO.
Given a ternary $C^*$-ring $X$, let $A$ be the closed linear span within $B(X,X)$ (bounded operators on $X$) of the set
of operators of the form
$$
  T_{y, z}:x\in X\mapsto  [x,y,z]\in X,
  $$
as $y$ and $z$ range in $X$.  It is easy to see that $A$ is a Banach algebra, and Zettl proves that $A$ is indeed a
$C^*$-algebra for a unique involution operation "$^*$" satisfying
$$
  T_{y, z}^* = T_{z, y}.
  $$
Given this, it is clear that an operator of the form $T_{y,y}$ is self-adjoint but the key question is whether or not
this is moreover positive.
If $X$ is a TRO, then $T_{y, y}\geq 0$, while in the anti-TRO case, one has that $T_{y, y}\leq 0$.
In other words, the positivity of $T_{y, y}$ is a signature of TRO's not shared by their anti-TRO cousins.
[1] Zettl, Heinrich, A characterization of ternary rings of
operators, Adv. Math. 48, 117-143
(1983). ZBL0517.46049.
