Here is a lower bound of $2^{\Omega(2^n)}$ for the number of unimodular triangulations.

Let me start with the triangulation described by Włodzimierz Holsztyński, which is indeed quite classical and I will call the standard triangulation of the $n$-cube. Its $n$-simplices have as vertices the $n!$ monotone paths of vertices from $(0,\dots,0)$ to $(1,\dots,1)$.

Observe that the standard triangulation $T$ has the following property: for each $2$-face of the cube, the two triangles in which $T$ divides that face have the same link (that is, they are joined to the same simplices). This implies we can perform a flip on that $2$-face: we can remove the two triangles in it together with all simplices containing them, and insert the other triangulation of the quadrilateral, with the same link as the old one had. (This is a higher dimensional analogue of changing the triangulation in a square pyramid, as described in Eppstein's answer)

Now, if we find $k$ of these $2$-faces in which we can perform flips independently (that is, if no two of the flips affect a common simplex) then we get $2^k$ triangulations of the cube. One way to guarantee that the flips are independent is to use $2$-faces at the same level, by which I mean the following: every $2$-face is defined by $n-2$ equalities of the form $x_i=\epsilon_i$, where $\epsilon_i\in \{0,1\}$. I say a $2$-face has level $l$ if exactly $l$ of the $\epsilon$'s are equal to $1$.

The number of $2$-faces at level $l$ is ${n \choose n-2}{n-2 \choose l}$ which, for $l\sim (n-2)/2$, grows as $2^n n^{3/2}$ (modulo a constant). Thus, we have at least $$ 2^{\Omega(2^n n^{3/2})} $$ unimodular triangulations of the $n$-cube.

It is also easy to give a crude upper bound that is not that far from the lower bound. Observe that a unimodular triangulation is a set of $n!$ $n$-simplices taken out from a total of (at most) $2^n \choose n+1$ $n$-simplices. Thus, the number of unimodular triangulations (or, of all triangulations, for that matter) is at most $$ {{2^n \choose n+1} \choose n^!} \le 2^{O(n^{n+2})} $$

Here is a lower bound of $2^{\Omega(2^n)}$ for the number of unimodular triangulations.

Let me start with the triangulation described by Włodzimierz Holsztyński, which is indeed quite classical and I will call the standard triangulation of the $n$-cube. Its $n$-simplices have as vertices the $n!$ monotone paths of vertices from $(0,\dots,0)$ to $(1,\dots,1)$.

Observe that the standard triangulation $T$ has the following property: for each $2$-face of the cube, the two triangles in which $T$ divides that face have the same link (that is, they are joined to the same simplices). This implies we can perform a flip on that $2$-face: we can remove the two triangles in it together with all simplices containing them, and insert the other triangulation of the quadrilateral, with the same link as the old one had. (This is a higher dimensional analogue of changing the triangulation in a square pyramid, as described in David Eppstein's or TMA's answers)

Now, if we find $k$ of these $2$-faces in which we can perform flips independently (that is, if no two of the flips affect a common simplex) then we get $2^k$ triangulations of the cube. One way to guarantee that the flips are independent is to use $2$-faces at the same level, by which I mean the following: every $2$-face is defined by $n-2$ equalities of the form $x_i=\epsilon_i$, where $\epsilon_i\in \{0,1\}$. I say a $2$-face has level $l$ if exactly $l$ of the $\epsilon$'s are equal to $1$.

The number of $2$-faces at level $l$ is ${n \choose n-2}{n-2 \choose l}$ which, for $l\sim (n-2)/2$, grows as $2^n n^{3/2}$ (modulo a constant). Thus, we have at least $$ 2^{\Omega(2^n n^{3/2})} $$ unimodular triangulations of the $n$-cube.

It is also easy to give a crude upper bound that is not that far from the lower bound. Observe that a unimodular triangulation is a set of $n!$ $n$-simplices taken out from a total of (at most) $2^n \choose n+1$ $n$-simplices. Thus, the number of unimodular triangulations (or, of all triangulations, for that matter) is at most $$ {{2^n \choose n+1} \choose n!} \le 2^{O(n^{n+2})} $$