For $n\ge 6$, it takes $n+1$ blocks - see part 2 for the construction. For $2\le n\le 5$, it takes $n+2$ blocks. The $n+2$-block shape that works:
Part 1
Label the blocks based on their centers, placed at lattice points. The shape that does it has blocks at $(-1,0,0,\dots)$, $(0,0,0,\dots)$, $(1,0,0,\dots)$, $(1,1,0,\dots)$, $(1,1,1,\dots)$, and so on until we fill in all of the coordinates with ones.
Why does this work? We prove it inductively, taking the $2$-dimensional L-shape as the base case and showing that an isometry (of any dimension) that fixes the shape must fix each individual block. Our shape is a "worm" with each block connected to two others except for the ends which are only connected to one block each. The two ends are also distinguishable; only one is part of a three-block straight line. Call that end the "head" and the end that isn't part of such a block the tail.
So then, suppose an isometry maps the shape to itself. Since numbers of connections stay the same, the ends must map to the ends. Since the ends are distinguishable, the head must map to the head and the tail must map to the tail. Cut off the tail block, reducing us to the shape with one less dimension. By the inductive hypothesis, the isometry fixes each block of the reduced shape. Add the fixed tail block back in, and the isometry fixes all of the blocks.
Finally, take the $n$-dimensional shape and restrict our isometry to an isometry of $\mathbb{R}^n$. The vector differences between centers of blocks span $\mathbb{R}^n$, so an isometry that fixes every block fixes a spanning set, and must fix all of $\mathbb{R}^n$. Done.
Part 2
While the above shape always works, it isn't necessarily minimal. To pin down the exact minimum, first note that any shape with $n$ or fewer blocks in it fits in $n-1$ or fewer dimensions; each new block is connected to a previous block, and those $\le n-1$ vectors don't span the full space. Then we just have to take an isometry that fixes all of the dimensions the shape is in, and reflects in one of the other dimensions. Therefore, the minimum is always at least $n+1$.
For $n\ge 6$, the minimum is equal to $n+1$. The seven blocks of the six-dimensional shape: $(0,0,0,0,0,0)$, $(1,0,0,0,0,0)$, $(0,1,0,0,0,0)$, $(0,1,1,0,0,0)$, $(0,0,0,1,0,0)$, $(0,0,0,1,1,0)$, $(0,0,0,1,1,1)$. We have a base block with three "tails" of different lengths coming from it, and that base block is the only block connected to more than two other blocks. Since the tails have different lengths, they're distinguishable and any isometry must fix each tail. Like our $n+2$ example, this uses all the dimensions, and thus any isometry that fixes all the blocks fixes everything.
For $n>6$, we extend this shape by adding more blocks to the longest tail, up to $(0,0,0,1,1,1,\dots,1)$. All the key properties remain, and the shape has no nontrivial isometries.
Part 3
That leaves $n\le 5$. To get a slightly better understanding, I'll turn to graph theory here; if we take the blocks as vertices and adjacent connections as edges, we have a graph. A shape of $n+1$ blocks that uses all dimensions must have its graph be a tree; the $n$ edges are linearly independent, and there are no cycles
Now, any graph automorphism of this tree induces an isometry of the shape. We will now show that any tree on three to six vertices has nontrivial automorphisms, and thus any $n$-dimensional shape of $n+1$ blocks has a nontrivial isometry for $2\le n\le 5$.
Three vertices: There's only one tree up to automorphism. Swapping the two ends is an automorphism.
Four vertices: There are two trees up to automorphism:
Both of them have nontrivial automorphisms; reverse the order for the first, shuffle the leaves for the second.
Five vertices: There are three trees up to automorphism:
All have nontrivial automorphisms, easily visualized.
Six vertices: There are six trees up to automorphism:
Once again, all have nontrivial automorphisms; swapping tails of equal length always works.
since you can reflect it along the vertical axis and get the same shape.
