I believe there is a regular non-sequential almost sequential space in ZFC+CH.
For $S,T\subseteq \omega$ let $S\subseteq^* T$ denote inclusion modulo finite sets i.e. $S\setminus T$ is finite. For $f,g:\omega\to\omega$ let $f\leq^* g$ denote dominance modulo finite sets i.e. $f(n)\leq g(n)$ except for finitely many $n.$ Greek letters will denote elements of $\omega_1.$
The construction will make use of an ultrafilter $\mathcal U$ on $\omega,$ a cofinal increasing $\omega_1$-sequence $S_\alpha$ in $(\mathcal U,\supseteq^*),$ and a cofinal increasing $\omega_1$ sequence $f_\alpha$ in $(\omega^\omega,\leq^*).$ So $\alpha<\beta$ implies $S_\alpha\supset^* S_\beta,$ and for every set $S$ there is $\alpha$ such that $S\supseteq^* S_\alpha$ or $\omega\setminus S\supseteq^* S_\alpha.$ And $\alpha<\beta$ also implies $f_\alpha\leq^*f_\beta,$ and for every $f$ there is $\alpha$ such that $f\leq^* f_\alpha.$ These are easy to construct under CH by transfinite induction. Specifically, we can take $S_\alpha$ to be a strictly $\subseteq^*$-increasing subsequence of the sets called $X_\alpha$ in the construction of a Ramsey ultrafilter in Jech's Set theory Theorem 7.8 (3rd ed), and take $f_\alpha(n)$ to be smallest integer with $|S_\alpha\cap\{0,1,\dots,f_\alpha(n)\}|=n.$
The ordinals $\omega$ and $\omega^+$ have the ordinal topology. Let $X$ be the topological space on the set $(\omega\times\omega^+)\cup\{*\}$ generated by open sets in $\omega\times\omega^+$ and the sets $U_{\alpha,n}$ defined for all $\alpha\in\omega_1$ and all integers $n$ by $$U_{\alpha,n}=\{*\}\cup\{(x,y)\mid x>n\text{ and either }x\in S_\alpha\text{ or }y\leq f_\alpha(x)\}.$$ The sets $U_{\alpha,n}$ are a neighborhood subbase of $\{*\}.$
$X$ is regular. It has a subbase of clopen sets.
$X$ is not sequential. The subspace $A=\omega\times\{\omega\}$ has $\{*\}$ as a limit point. This is because any finite intersection $U_{\alpha_1,n_1}\cap \dots\cap U_{\alpha_k,n_k}\cap A$ is just $S_{\max(\alpha_i)}\times\{\omega\}$ minus a finite set, and is therefore non-empty. Suppose for contradiction that a sequence $(x_n,\omega)$ converges to $*.$ Split $\{x_n\}$ into two infinite sets. One of these sets, call it $S,$ is not in $\mathcal U.$ There is $\alpha$ such that $\omega\setminus S\supseteq^* S_\alpha.$ The corresponding subsequence therefore lies outside $U_{\alpha,n}$ for sufficiently large $n.$
$X$ is almost sequential. Each $x\neq *$ lies in the dense sequential subspace $X\setminus\{*\}.$ The point $*$ lies in the subspace $A=(\omega\times\omega)\cup \{*\}$ which is clearly dense, and it remains to show that it is sequential. The only problem is at $*.$ Accordingly, consider a set $C\subseteq A\setminus\{*\}$ whose closure is $C\cup \{*\},$ and we need to exhibit a sequence converging to $*.$ Define $C_\gamma=\{(x,y)\in C\mid x\in S_\gamma\}.$
First consider the case that $*$ is a limit point of $C_\gamma$ for every $\gamma.$ Let $D=\{x\mid \exists y.(x,y)\in C\}.$ Pick any function $f:\omega\to\omega$ with $(x,f(x))\in C$ for each $x\in D.$ Pick $\alpha$ such that $f\leq^* f_\alpha.$ Take any strictly increasing sequence $x_n$ in $D\cap S_\alpha.$ Consider an arbitrary $U_{\beta,N}.$ We either have $\beta\leq\alpha$ giving $x_n\in S_\beta$ eventually, or $\beta\geq\alpha$ giving $f(x_n)\leq f_\beta(x_n)$ eventually. This proves that $(x_n,f(x_n))$ converges to $*.$
Now assume that $*$ is not a limit point of $C_\gamma,$ for some $\gamma.$ Take $\gamma$ to be minimal. So $*$ is a limit point of $C_{\beta}$ if and only if $\beta<\gamma.$ Since $0\neq\gamma<\omega_1$ there is a countable sequence $\beta_n$ with $\sup(\beta_n^+)=\gamma.$ For each $n,N$ the set $C_{\beta_n}\setminus C_\gamma$ must intersect the neighborhood $U_{\gamma,N},$ which means there are $(x,y)\in C_{\beta_n}\setminus C_\gamma$ with $x>N$ and $y\leq f_\gamma(x).$ Therefore $C_{\beta_1}\cap\dots\cap C_{\beta_n}\setminus C_\gamma$ (which might be smaller than $C_{\beta_n}\setminus C_\gamma$ at a finite number of $x$-coordinates) contains some $(x_n,y_n)$ with $x>N$ and $y\leq f_\gamma(x).$ We can pick such a choice of $(x_n,y_n)$ for each $n,$ using $N$ to ensure that $x_n$ is strictly increasing. This construction ensures that for each $\beta<\gamma$ the sequence $(x_n,y_n)$ eventually lies in $C_{\beta},$ and $y_n\leq f_\gamma(x_n)$ for all $n.$ So for each $\beta,n$ the sequence $(x_n,y_n)$ eventually lies in $U_{\beta,n}.$ This proves that $(x_n,y_n)$ converges to $*.$