Any compactly generated presentable stable $\infty$-category $C$ is known to be dualizable (with respect to Lurie's tensor product), so there is a coevaluation map:
$$Sp \to C \otimes C^{dual}.$$
Can one describe this map (or the image of the sphere spectrum) more concretely in terms of the compact generators of $C$?
Recall that, for $A,B$ stable presentable $\infty$-categories, we can compute the tensor and internal hom using the equivalences $A\otimes B \simeq \operatorname{Fun}^\mathrm{lim}(A^\mathrm{op},B)$ and $\operatorname{hom}(A,B) \simeq \operatorname{Fun}^\mathrm{L}(A,B)$. If $A$ is moreover compactly generated, we also have $\operatorname{Fun}^\mathrm{L}(A,B) \simeq \operatorname{Fun}^\mathrm{ex}(A^\omega, B)$.
Thus, for $C$ a stable compactly-generated $\infty$-category, a colimit-preserving functor $\mathrm{Spt} \to C \otimes C^\vee$ is the same thing as an object of $C \otimes C^\vee \simeq \operatorname{Fun}^\mathrm{lim}(C^\mathrm{op}, \operatorname{Fun}^\mathrm{ex}(C^\omega,\mathrm{Spt}))$. So, what are some good functors $C^\mathrm{op} \times C^\omega \to\mathrm{Spt}$? One reliable choice is the (restricted) mapping spectrum functor $(x,y)\mapsto \operatorname{map}(x,y)$. This is a sensible choice, since the functor $C^\mathrm{op} \to \operatorname{Fun}(C^\omega,\mathrm{Spt})$ given by $x\mapsto \operatorname{map}(x,{-})$ preserves limits and factors through the full subcategory $\operatorname{Fun}^\mathrm{ex}(C^\omega,\mathrm{Spt})$ of the target. And indeed, one can check that this, together with the evaluation functor $C^\vee \otimes C \to \mathrm{Spt}$ defined by $(F, x) \mapsto F(x)$, satisfies the triangle identities for dual objects.
Alternatively, the object of $C\otimes C^\vee$ that we are interested in can be described as the coend $\int^{x\in C^\omega} x\otimes \operatorname{map}(x,{-})$, which is might align better with one's linear-algebraic intuitions.
Brian's answer is really what it is, but let me give a second perspective on it.
In general, if $x$ is dualizable, $x\otimes x^\vee \simeq \hom(x,x)$, where $\hom$ is the internal hom, and essentially by design of this identification, the co-unit $1\to x\otimes x^\vee$ coincides with $1\to \hom(x,x)$ picking out the identity.
In our case, the internal hom has a description independent of duals : it's $Fun^L(C,C)$, left adjoint functors $C\to C$.
So really, we need to trace through the identification $C\otimes C^\vee \simeq Fun^L(C,C)$ and see who corresponds to the identity.
Now $C^\vee = Fun^L(C,Sp)= Fun^{ex}(C^\omega, Sp)$ and the canonical map $(C^\omega)^{op}\to C^\vee$ that identifies the latter with the ind-completion of the former corresponds simply to $x\mapsto Map(-,x)$ ($Map$ here is the mapping spectrum)
Finally, the identification $Ind(C^\omega \otimes (C^\omega)^{op}) \simeq C\otimes Ind((C^\omega)^{op})\simeq C\otimes Fun^{ex}(C^\omega, Sp)= Fun^L(C,C) $ sends $x\otimes y \mapsto x\otimes Map(y,-)$ where the latter can be interpreted as a formal tensor product, or as the tensoring of $x$ by some spectrum.
So in a sense you're expecting a description of $id_C$ in terms of $x\otimes Map(y,-)$'s. A priori it's not super easy, but if you think in terms of Yoneda, the coend mentioned by Brian comes out. Namely, for another functor $F$, one finds $Map_{Fun^L(C,C)}(id_C, F)= Map_{Fun^{ex}(C^\omega, C)}(i,F_{\mid C^\omega}) = \int_{x\in C^\omega}Map_C(x, F(x))$, using the description of natural transformation spaces/spectra as ends.
And then you observe that, by a spectral Yoneda lemma, $Map_C(x,F(x)) \simeq Map_{Fun^L(C,C)}(Map(x,-), Map(x,F(-))\simeq Map_{Fun^L(C,C)}(x\otimes Map(x,-), F(-))$.
So you can rewrite your end as $\int_{x\in C^\omega}Map_{Fun^L(C,C)}(x\otimes Map(x,-), F) \simeq Map(\int^{x\in C^\omega} x\otimes Map(x,-), F)$ where the last bit is some co/end calculus. By one (final!) application of the Yoneda lemma, you find that $id_C\simeq \int^{x\in C^\omega} x\otimes Map(x,-)$, but now if you treat $Map(x,-)$ as "$x$ but in $(C^\omega)^{op}$, this gives you a well-defined expression in $C\otimes C^\vee$, which is $\int^{x\in C^\omega}x\otimes x^\vee$ (where the latter $^\vee$ should not be understood as a symmetric monoidal dual, just a formal dual).
