Let $\Gamma$ be a discrete group acting on an infinite-dimensional Banach space $X$ by linear isometries.
Is there a probability measure (non-atomic, not supported on a finite dimensional subspace) on Borel subsets of $B_X$, the unit ball of $X$, that is invariant under this action? If not in general then under what assumptions?
The answer is: yes iff $X$ contains a finite dimensional invariant subspace of arbitrary large dimension (as you see, it has really nothing to do with the Banach structure of $X$ only with its quality as a $\Gamma$-representation). Equivalently, the answer is: yes iff $X_0$ (defined below) is of infinite dimension.
First example (which you will not like): take a trivial representation of $\Gamma$ on $X$ and any probability measure supported on $B_X$ (but not on a finite dimensional subspace).
You want to disregard the first example as non-ergodic.
Second example (very explicit): Take $\Gamma= \mathbb Z$ and let it act by an irrational rotation (by $\sqrt{2}$, as I said I'll be explicit) on $S^1$. Consider $X=L^2(S^1)$. In $B_X$ take the subset $Y$ consisting of all characteristic functions of half circles in $S^1$ (normalized). $Y$ is an $S^1$ equivariant copy of $S^1$ in $B_X$ and we can endow it with the Haar measure of $S^1$. This measure satisfies everything you wished for.
The second example is typical in the sense that the $\Gamma$-invariant measure is actually invariant under a compact group enveloping $\Gamma$.
I will explain below that if $X$ has no finite dimensional invariant subspaces (other than $\{0\}$) then the only invariant probability measure on $X$ is the Dirac measure at $0$ (for $X$ Hilbert this is Lemma 5.6 in "Amenable Invariant Random Subgroups" by Bader*-Duchesne-Lecureux). More generally, let $X_0$ be the closure in $X$ of the vector space generated by all invariant finite dimensional subspaces. Then every invariant probability measure on $X$ is supported on $X_0$.
It follows that if $X_0$ is finite dimensional then every invariant measure is indeed supported on finite dimensional invariant subspace. Note that if $X_0$ is infinite dimensional then it contains a sequence of finite dimensional invariant subspaces of arbitrary large dimensions and summing up the Lebesgue measures of the corresponding spheres (with coefficients so that the total mass is 1) gives an invariant probability measure which is not supported on a finite dimensional invariant subspace.
In what follows I will discuss $\Gamma$-invariant measures on the Banach space $X$ (not necessarily on the unit ball). I will assume that $X$ is separable, which probably is enough for you (e.g if $\Gamma$ is countable, or more generally, if $\Gamma$ is locally compact second countable and it acts strongly continuously on $X$). Fix a probability measure $\nu$ on the orbit space $X_0/G$, where $G$ is the Bohr compactification of $\Gamma$ (recall that the $\Gamma$ action on $X_0$ extends to $G$) and set $$\mu=\int_{X_0/G} (\text{The normalized Haar measure on the orbit})~ d\nu(\text{orbit}).$$ Then $\mu$ is clearly a $\Gamma$-invariant measure and it is supported on $X_0$. Conversely, every $\Gamma$-invariant measure is given in this form. In particular, every ergodic $\Gamma$-measure is supported (on $X_0$ and) on a unique $G$-orbit, and it is the Haar measure on it. Note that by ergodic decomposition it is enough to prove the last statement.
Claim: let $\Gamma$ be a group acting isometrically on a complete, separable metric space $Y$ preserving a fully supported, ergodic probability measure $\mu$. Then there is a $\Gamma$-equivariant homeomorphism $Y\simeq G/L$, where $G$ is the Bohr compactification of $\Gamma$ and $L<G$ a closed subgroup, and under this identification $\mu$ corresponds to the Haar measure.
The essential part of the claim is the compactness of $Y$, which follows by completeness from total-boundedness, which is easy (generalize the standard claim that $Y$ is bounded, which is proven by taking a compact set of more than half the weight). Then the isometry group of $Y$ is compact by Arzela-Ascoli and the $\Gamma$ action extends to $G$. By ergodicity (and the fact that $Y/G$ is nice) we get that $\mu$ is supported on a unique orbit and everything else follows easily.
Now, for the statement we had regarding ergodic measures on $X$, take $Y$ to be the support of an ergodic measure on $X$, which I assume to be total. Observe that the action of $G$ on $Y$ extends to a linear isometric action on $X$. Call Peter-Weyl to see that $X=X_0$.
$*$ Hi.
