A deterministic function $f(X)$ of a random variable $X$ never has greater entropy than the original random variable $X$.
(Side note: If you're not familiar with the technical terms random variable and expectation, or the definition of entropy in terms of them, I recommend you read up on some elementary probability theory first—the rest of this post will not make sense without these keywords! In the sequel, ‘entropy’ can be taken to mean Shannon entropy or min-entropy—both work, and coincide on uniform distributions, but cryptography works primarily in min-entropy. See this answer for a brief discussion of what entropy means from a crypto perspective.)
At best, if $X$ is a random $k$-bit string with uniform distribution and thus $k$ bits of entropy, the maximum possible, and if $f\colon \{0,1\}^k \to \{0,1\}^k$ is a permutation (a.k.a. bijection), then $f(X)$ also has $k$ bits of entropy. If $f$ is not a permutation, then $f(X)$ has strictly less than $k$ bits of entropy—although perhaps not much less.
In this case, $f$ is SHA-256, which we might model by a uniform random choice of map $F\colon \{0,1\}^k \to \{0,1\}^k$. There are almost certainly fewer than $2^k$ distinct outputs, because very few such possible values of $F$ are permutations. Specifically, fix an output $y \in \{0,1\}^k$; for each input $x \in \{0,1\}^k$, we have $\Pr[F(x) = y] = 1/2^k$, and $F(x)$ is an independent random variable for every $x$, so
\begin{align*}
\Pr&[\exists x. F(x) = y]
= 1 - \Pr[\forall x. F(x) \ne y] \\
&= 1 - \Pr[F(0) \ne y]\,\Pr[F(1) \ne y]\cdots\Pr[F(2^k - 1) \ne y] \\
&= 1 - (1 - 1/2^k)^{2^k}.
\end{align*}
which, by linearity of expectations, is also the expected fraction of distinct outputs. As $k \to \infty$, this converges to $1 - e^{-1} \approx 0.632$.
Thus, over a uniform random choice of $F$, the expected entropy of $F(X)$, or $\mathbb E_F\bigl[H[F(X)]\bigr]$, is about one bit less than the entropy of $X$, or $H[X]$, where $H$ is the entropy operator.
What about iterating SHA-256? Do we lose a bit of entropy every time, so that $\operatorname{SHA-256}^{256}$ has zero entropy? No. If we independently chose another function $G\colon \{0,1\}^k \to \{0,1\}^k$, then over a uniform random choice of $F$ and $G$, the expected entropy of $F(G(X))$ might be about two bits less than the entropy of $X$. But that's a bad model for iterating SHA-256, because we are restricted to the case $F = G$, in which case $F$ and $G$ are as far from independent as possible.
Rather, there's a good chance that for any fixed function $F$, there are many cycles on which $F$ is a permutation, and restricted to which $F$ thus preserves entropy. There's usually one main big cycle, which lends itself to a drawing of a giant hairy rho: $\rho$. In the limit as $\ell \to \infty$, $F^\ell(X)$ maps $X$ to an independent uniform random point on one of the cycles, the choice of cycle being weighted by the number of points on and leading to that cycle.
But in practical terms for cryptography engineering? Usually we treat $\operatorname{SHA-256}(X)$ as a uniform random variable independent of everything else in the system (except where length extension attacks might be relevant) with entropy equal to that of $X$, or 256, whichever is smaller.
