KL divergence is a natural way to measure the difference between two probability distributions. The entropy $H(p)$ of a distribution $p$ gives the minimum possible number of bits per message that would be needed (on average) to losslessly encode events drawn from $p$. Achieving this bound would require using an optimal code designed for $p$, which assigns shorter code words to higher probability events. $D_{KL}(p \parallel q)$ can be interpreted as the expected number of extra bits per message needed to encode events drawn from true distribution $p$, if using an optimal code for distribution $q$ rather than $p$. It has some nice properties for comparing distributions. For example, if $p$ and $q$ are equal, then the KL divergence is 0.
The cross entropy $H(p, q)$ can be interpreted as the number of bits per message needed (on average) to encode events drawn from true distribution $p$, if using an optimal code for distribution $q$. Note the difference: $D_{KL}(p \parallel q)$ measures the average number of extra bits per message, whereas $H(p, q)$ measures the average number of total bits per message. It's true that, for fixed $p$, $H(p, q)$ will grow as $q$ becomes increasingly different from $p$. But, if $p$ isn't held fixed, it's hard to interpret $H(p, q)$ as an absolute measure of the difference, because it grows with the entropy of $p$.
KL divergence and cross entropy are related as:
$$D_{KL}(p \parallel q) = H(p, q) - H(p)$$
We can see from this expression that, when $p$ and $q$ are equal, the cross entropy is not zero; rather, it's equal to the entropy of $p$.
Cross entropy commonly shows up in loss functions in machine learning. In many of these situations, $p$ is treated as the 'true' distribution, and $q$ as the model that we're trying to optimize. For example, in classification problems, the commonly used cross entropy loss (aka log loss), measures the cross entropy between the empirical distribution of the labels (given the inputs) and the distribution predicted by the classifier. The empirical distribution for each data point simply assigns probability 1 to the class of that data point, and 0 to all other classes. Side note: The cross entropy in this case turns out to be proportional to the negative log likelihood, so minimizing it is equivalent maximizing the likelihood.
Note that $p$ (the empirical distribution in this example) is fixed. So, it would be equivalent to say that we're minimizing the KL divergence between the empirical distribution and the predicted distribution. As we can see in the expression above, the two are related by the additive term $H(p)$ (the entropy of the empirical distribution). Because $p$ is fixed, $H(p)$ doesn't change with the parameters of the model, and can be disregarded in the loss function. We might still want to talk about the KL divergence for theoretical/philosophical reasons but, in this case, they're equivalent from the perspective of solving the optimization problem. This may not be true for other uses of cross entropy and KL divergence, where $p$ might vary.
t-SNE fits a distribution $p$ in the input space. Each data point is mapped into the embedding space, where corresponding distribution $q$ is fit. The algorithm attempts to adjust the embedding to minimize $D_{KL}(p \parallel q)$. As above, $p$ is held fixed. So, from the perspective of the optimization problem, minimizing the KL divergence and minimizing the cross entropy are equivalent. Indeed, van der Maaten and Hinton (2008) say in section 2: "A natural measure of the faithfulness with which $q_{j \mid i}$ models $p_{j \mid i}$ is the Kullback-Leibler divergence (which is in this case equal to the cross-entropy up to an additive constant)."
van der Maaten and Hinton (2008). Visualizing data using t-SNE.
Let $q$ be the density of your true data-generating process and $f_\theta$ be your model-density.
Then $$KL(q||f_\theta) = \int q(x) log\left(\frac{q(x)}{f_\theta(x)}\right)dx = -\int q(x) \log(f_\theta(x))dx + \int q(x) \log(q(x)) dx$$
The first term is the Cross Entropy $H(q, f_\theta)$ and the second term is the (differential) entropy $H(q)$. Note that the second term does NOT depend on $\theta$ and therefore you cannot influence it anyway. Therfore minimizing either Cross-Entropy or KL-divergence is equivalent.
Without looking at the formula you can understand it the following informal way (if you assume a discrete distribution). The entropy $H(q)$ encodes how many bits you need if you encode the signal that comes from the distribution $q$ in an optimal way. The Cross-Entropy $H(q, f_\theta)$ encodes how many bits on average you would need when you encoded the singal that comes from a distribution $q$ using the optimal coding scheme for $f_\theta$. This decomposes into the Entropy $H(q)$ + $KL(q||f_\theta)$. The KL-divergence therefore measures how many additional bits you need if you use an optimal coding scheme for distribution $f_\theta$ (i.e. you assume your data comes from $f_\theta$ while it is actually generated from $q$). This also explains why it has to be positive. You cannot be better than the optimal coding scheme that yields the average bit-length $H(q)$.
This illustrates in an informal way why minimizing KL-divergence is equivalent to minimizing CE: By minimzing how many more bits you need than the optimal coding scheme (on average) you of course also minimize the total amount of bits you need (on average)
The following post illustrates the idea with the optimal coding scheme: Qualitively what is Cross Entropy
Best Answer
It seems the answer is no. According to your expression:
$$D_{KL} (u \parallel p) = -\log n - \frac{1}{n} \sum_x \log p(x)$$
The entropy of $p$ is
$$H(p) = -\sum_x p(x) \log p(x)$$
But, there's no way to recover this from the first expression. Given $D_{KL}(u \parallel p)$ (and no knowledge of $p$), the closest we could come is to add $\log n$ then multiply by $n$ to obtain $-\sum_x \log p(x)$. But, this sum collapses everything into a single value, and we cant recover the individual probabilities needed to compute the entropy.
But, what you want to do is possible if the KL divergence is computed the other way around:
$$D_{KL}(p \parallel u) = \sum_x p(x) \log p(x) - \sum_x p(x) \log u(x)$$
The first term is the negative entropy:
$$= -H(p) - \sum_x p(x) \log u(x)$$
Plug in the uniform probabilities:
$$= -H(p) - \sum_x p(x) \log \frac{1}{n}$$
Simplify the last term (noting that the distribution sums to one):
$$= -H(p) + \log n$$
Therefore:
$$H(p) = \log n - D_{KL}(p \parallel u)$$